Areas of Plane Figures: Concepts and Theorems
Area of Polygonal Regions: Basic Concepts
The concept of area is fundamental in geometry and measurement. It quantifies the extent of the two-dimensional space occupied by a closed planar figure. Every simple closed plane figure, such as a polygon, divides the plane into three distinct parts:
- The interior of the figure (the region strictly inside the boundary).
- The boundary of the figure (the line segments that form the perimeter).
- The exterior of the figure (the region strictly outside the boundary).
Planar Region and Area
The planar region corresponding to a closed figure is the union of its boundary and its interior. It represents the entire flat surface covered by the figure.
The area of a figure is a positive numerical value assigned to its planar region. It provides a measure of the "size" of the surface enclosed by the figure's boundary. The area allows us to compare the sizes of different two-dimensional shapes.
Properties of Area
The concept of area is based on certain fundamental properties or axioms:
-
Congruence Property: If two planar figures are congruent, then they have the same area.
Mathematically, if Figure A $\cong$ Figure B, then $\text{Area(A)} = \text{Area(B)}$.
Important Note: The converse of this property is not true. Two figures can have the same area without being congruent. For example, a square with side length 4 cm has an area of $4^2 = 16 \text{ cm}^2$. A rectangle with dimensions 8 cm by 2 cm also has an area of $8 \times 2 = 16 \text{ cm}^2$. However, a square and a non-square rectangle are not congruent figures.
-
Additive Property: If a planar region (let's call it T) is composed of two or more non-overlapping planar regions (say P and Q), then the area of the larger region T is the sum of the areas of the smaller regions P and Q.
Mathematically, if region T is the union of non-overlapping regions P and Q, then $\text{Area(T)} = \text{Area(P)} + \text{Area(Q)}$.
This property is useful when dealing with composite shapes or when a complex shape can be divided into simpler shapes.
-
Unit Property: Area is always measured in square units. The standard unit of area in the International System of Units (SI) is the square meter ($\text{m}^2$). Other common units include square centimeters ($\text{cm}^2$), square kilometers ($\text{km}^2$), square inches ($\text{in}^2$), square feet ($\text{ft}^2$), etc. The choice of unit depends on the scale of the figure being measured.
For example, the area of a square with side length 1 unit is defined as 1 square unit.
In mathematical notation, the area of a figure is often denoted by writing "ar" before the name of the figure or the vertices defining it. For instance, the area of a triangle with vertices A, B, and C is denoted as $\text{ar}(\triangle ABC)$. Similarly, the area of a quadrilateral PQRS is denoted as $\text{ar}(PQRS)$.
Figures on the Same Base and Between the Same Parallels
This concept is fundamental in the study of areas of parallelograms and triangles. It establishes a specific geometric configuration that leads to important theorems about equal areas.
Two or more geometric figures (polygons) are said to be on the same base and between the same parallels if they satisfy the following two conditions simultaneously:
- They share a common side, which is referred to as the 'base'.
- The vertex or vertices (or the entire side, in the case of the second base of a parallelogram or the third vertex of a triangle) of each figure, which are opposite to the common base, must lie on a single straight line that is parallel to the line containing the common base.
Illustrations of Figures on the Same Base and Between the Same Parallels
1. Parallelograms on the same base and between the same parallels:
Consider two parallelograms ABCD and EBCF.
In the figure:
- Both parallelogram ABCD and parallelogram EBCF share the side $\overline{BC}$ as their common base.
- The vertices opposite to the base $\overline{BC}$ are A and D for parallelogram ABCD, and E and F for parallelogram EBCF.
- These vertices (A, D, E, F) all lie on the same straight line (denoted as L1 in the figure).
- The line L1 containing A, D, E, F is parallel to the line L2 containing the common base $\overline{BC}$.
Therefore, parallelogram ABCD and parallelogram EBCF are on the same base $\overline{BC}$ and between the same parallels (the line through A, D, E, F and the line through B, C).
2. Triangles on the same base and between the same parallels:
Consider two triangles $\triangle ABC$ and $\triangle DBC$.
In the figure:
- Both $\triangle ABC$ and $\triangle DBC$ share the side $\overline{BC}$ as their common base.
- The vertices opposite to the base $\overline{BC}$ are A for $\triangle ABC$ and D for $\triangle DBC$.
- These vertices (A and D) lie on the same straight line (denoted as L1 in the figure).
- The line L1 containing A and D is parallel to the line L2 containing the common base $\overline{BC}$.
Therefore, $\triangle ABC$ and $\triangle DBC$ are on the same base $\overline{BC}$ and between the same parallels (the line through A, D and the line through B, C).
3. Triangle and Parallelogram on the same base and between the same parallels:
Consider a parallelogram ABCD and a triangle $\triangle EBC$.
In the figure:
- Both parallelogram ABCD and $\triangle EBC$ share the side $\overline{BC}$ as their common base.
- The vertices opposite to the base $\overline{BC}$ are A and D for parallelogram ABCD, and E for $\triangle EBC$.
- These vertices (A, D, E) all lie on the same straight line (denoted as L1 in the figure).
- The line L1 containing A, D, E is parallel to the line L2 containing the common base $\overline{BC}$.
Therefore, parallelogram ABCD and $\triangle EBC$ are on the same base $\overline{BC}$ and between the same parallels (the line through A, D, E and the line through B, C).
This configuration is the prerequisite for several area-related theorems, most notably that figures configured this way have specific relationships between their areas.
Parallelograms on the Same Base and Between the Same Parallels (Theorem)
This theorem establishes a fundamental relationship between the areas of parallelograms that share a specific relative position.
Statement
Parallelograms on the same base and between the same parallels are equal in area.
Proof
Given:
Two parallelograms ABCD and EBCF.
They share the same base $\overline{BC}$.
They lie between the same parallel lines, namely the line containing $\overline{BC}$ and the line containing vertices A, D, E, and F (let's call this line AF, so $AF \parallel BC$).
To Prove:
The area of parallelogram ABCD is equal to the area of parallelogram EBCF, i.e., $ar(ABCD) = ar(EBCF)$.
Proof:
Consider the two triangles $\triangle ABE$ and $\triangle DCF$.
Since ABCD is a parallelogram, its opposite sides are parallel and equal:
$\overline{AB} \parallel \overline{DC}$
(Opposite sides of parallelogram ABCD) ... (1)
AB = DC
(Opposite sides of parallelogram ABCD) ... (2)
Since EBCF is a parallelogram, its opposite sides are parallel and equal:
$\overline{EB} \parallel \overline{FC}$
(Opposite sides of parallelogram EBCF) ... (3)
BE = CF
(Opposite sides of parallelogram EBCF) ... (4)
Now, consider the lines AF and BC as parallel lines intersected by transversals.
Consider transversal $\overline{AF}$ intersecting parallel lines $\overline{AB}$ and $\overline{DC}$. The corresponding angles are equal:
$\angle EAB = \angle FDC$
(Corresponding angles, $\overline{AB} \parallel \overline{DC}$ and $\overline{AF}$ transversal)
(Note: $\angle EAB$ is $\angle FAB$ at vertex A, and $\angle FDC$ is $\angle ADC$ at vertex D, if E, A, D, F are collinear in that order).
Consider transversal $\overline{AF}$ intersecting parallel lines $\overline{EB}$ and $\overline{FC}$. The corresponding angles are equal:
$\angle AEB = \angle DFC$
(Corresponding angles, $\overline{EB} \parallel \overline{FC}$ and $\overline{AF}$ transversal)
Now, in $\triangle ABE$ and $\triangle DCF$:
$\angle EAB = \angle FDC$
(Proved above)
$\angle AEB = \angle DFC$
(Proved above)
AB = DC
(From 2, opposite sides of parallelogram ABCD)
The side $\overline{AB}$ is opposite to $\angle AEB$ in $\triangle ABE$, and the side $\overline{DC}$ is opposite to $\angle DFC$ in $\triangle DCF$. Since $\angle AEB = \angle DFC$ and $AB = DC$, by the AAS (Angle-Angle-Side) congruence criterion, we have:
$\triangle ABE \cong \triangle DCF$
Since congruent triangles have equal areas, we have:
ar($\triangle ABE$) = ar($\triangle DCF$)
... (5)
Now, let's consider the area of the parallelograms. From the figure, we can see that the quadrilateral EBCD is common to both parallelograms in terms of area composition (regardless of the order of A, D, E, F on the top line, the net result of area addition/subtraction holds).
The area of parallelogram ABCD is the area of the region formed by $\triangle ABE$ and quadrilateral EBCD:
ar(ABCD) = ar($\triangle ABE$) + ar(EBCD)
The area of parallelogram EBCF is the area of the region formed by $\triangle DCF$ and quadrilateral EBCD:
ar(EBCF) = ar($\triangle DCF$) + ar(EBCD)
Using equation (5), substitute $ar(\triangle DCF)$ with $ar(\triangle ABE)$ in the second equation:
ar(EBCF) = ar($\triangle ABE$) + ar(EBCD)
Comparing this with the expression for $ar(ABCD)$, we get:
ar(ABCD) = ar(EBCF)
Thus, parallelograms on the same base and between the same parallels are equal in area.
Corollary: Area of a Parallelogram Formula
A direct consequence of this theorem is the well-known formula for the area of a parallelogram.
Statement: The area of a parallelogram is equal to the product of its base and the corresponding height.
Formula: $\text{Area of Parallelogram} = \text{Base} \times \text{Height}$
Derivation of the Formula
Consider a parallelogram ABCD with base $\overline{BC}$. Let $h$ be the height corresponding to the base $\overline{BC}$, which is the perpendicular distance between the parallel lines $\overline{BC}$ and $\overline{AD}$.
Construct a rectangle EBCF such that E and F lie on the line $\overline{AD}$ (or its extension) and $\overline{EB}$ and $\overline{FC}$ are perpendiculars to $\overline{BC}$. Thus, $\overline{EB}$ and $\overline{FC}$ are the heights, and $EB = FC = h$.
(Note: The provided image shows base AB. Let's adjust the derivation to match the standard base $\times$ height formula with base AB as in the image).
Consider a parallelogram ABCD with base $\overline{AB}$. Let $h$ be the height corresponding to the base $\overline{AB}$, which is the perpendicular distance between the parallel lines $\overline{AB}$ and $\overline{DC}$. Draw $\overline{DM} \perp \overline{AB}$ (extended if necessary) and $\overline{CN} \perp \overline{AB}$ (extended if necessary). Then $DM = CN = h$.
Construct a rectangle ABNM on the base $\overline{AB}$ such that M and N lie on the line $\overline{DC}$ (or its extension) and $\overline{AM}$ and $\overline{BN}$ are perpendiculars to $\overline{AB}$. Thus, $\overline{AM}$ and $\overline{BN}$ are the heights, and $AM = BN = h$.
Parallelogram ABCD and rectangle ABNM are on the same base $\overline{AB}$ and between the same parallel lines ($\overline{AB}$ and $\overline{DC}$).
By the theorem: Parallelograms on the same base and between the same parallels are equal in area.
ar(ABCD) = ar(ABNM)
The area of a rectangle is the product of its base and height.
ar(ABNM) = Base $\times$ Height = $AB \times BN = AB \times h$
Substituting this into the previous equation:
ar(ABCD) = $AB \times h$
Here, AB is the length of the base, and $h$ is the corresponding height. This proves that the area of a parallelogram is equal to the product of its base and the corresponding height.
Triangles on the Same Base and Between the Same Parallels (Theorem)
This theorem is analogous to the theorem about parallelograms and relates the areas of triangles that share a common base and are situated between the same pair of parallel lines.
Statement
Two triangles on the same base (or on equal bases) and lying between the same parallels are equal in area.
Proof
Given:
Two triangles, $\triangle ABC$ and $\triangle DBC$.
They share the same base $\overline{BC}$.
They lie between the same parallel lines, namely the line containing $\overline{BC}$ and the line containing vertices A and D (let's call this line AD, so $AD \parallel BC$).
To Prove:
The area of $\triangle ABC$ is equal to the area of $\triangle DBC$, i.e., $ar(\triangle ABC) = ar(\triangle DBC)$.
Construction:
Through vertex A, draw a line $\overline{AE}$ parallel to $\overline{BC}$, intersecting the line through C parallel to AB at E. (This constructs parallelogram ABCE).
Through vertex D, draw a line $\overline{DF}$ parallel to $\overline{BC}$, intersecting the line through C parallel to DB at F. (This constructs parallelogram DBCF).
(Note: Since A and D lie on the line parallel to BC, drawing lines parallel to BC through A and D is redundant; the construction aims to complete parallelograms related to the triangles and lying between the given parallels).
A simpler construction is: Draw a line through A parallel to $\overline{BC}$ and a line through D parallel to $\overline{BC}$. These lines must coincide with the line $\overline{AD}$ since A and D are already on a line parallel to $\overline{BC}$. Draw a line through B parallel to $\overline{AC}$, intersecting the line $\overline{AD}$ (extended if necessary) at E. Quadrilateral ABEC is a parallelogram.
Draw a line through B parallel to $\overline{DC}$, intersecting the line $\overline{AD}$ (extended if necessary) at F. Quadrilateral DBCF is a parallelogram.
Let's use the construction described in the provided image caption for consistency, which seems to involve completing parallelograms by drawing lines parallel to the non-base sides through the vertices opposite the base.
Construction based on image: Through C, draw $\overline{CE} \parallel \overline{BA}$ to meet the line AD extended at E. Through B, draw $\overline{BF} \parallel \overline{CD}$ to meet the line AD extended at F.
Proof (based on image construction):
Consider the quadrilateral ABCE. We are given $AD \parallel BC$, so $AE \parallel BC$. By construction, $CE \parallel BA$. Since both pairs of opposite sides are parallel, ABCE is a parallelogram.
The diagonal $\overline{AC}$ divides the parallelogram ABCE into two congruent triangles, $\triangle ABC$ and $\triangle ECA$. Therefore, their areas are equal, and each is half the area of the parallelogram.
ar($\triangle ABC$) = $\frac{1}{2}$ ar(parallelogram ABCE)
... (1)
Now consider the quadrilateral DBCF. We are given $AD \parallel BC$, so $DF \parallel BC$. By construction, $BF \parallel CD$. Since both pairs of opposite sides are parallel, DBCF is a parallelogram.
The diagonal $\overline{BD}$ divides the parallelogram DBCF into two congruent triangles, $\triangle DBC$ and $\triangle BFD$. Therefore, their areas are equal, and each is half the area of the parallelogram.
ar($\triangle DBC$) = $\frac{1}{2}$ ar(parallelogram DBCF)
... (2)
Now observe parallelograms ABCE and DBCF. They are both on the same base $\overline{BC}$. They lie between the same parallel lines, $\overline{BC}$ and the line $\overline{EF}$ (which is the line containing AD, extended). By the theorem "Parallelograms on the same base and between the same parallels are equal in area" (Section I3):
ar(parallelogram ABCE) = ar(parallelogram DBCF)
... (3)
Substitute equation (3) into equation (1):
ar($\triangle ABC$) = $\frac{1}{2}$ ar(parallelogram DBCF)
... (4)
Comparing equation (4) and equation (2), we see that both $ar(\triangle ABC)$ and $ar(\triangle DBC)$ are equal to $\frac{1}{2}$ ar(parallelogram DBCF).
Therefore:
ar($\triangle ABC$) = ar($\triangle DBC$)
Hence, triangles on the same base and between the same parallels are equal in area.
(Note: This theorem also holds for triangles on equal bases and between the same parallels. If two triangles have bases of equal length on the same line, and their opposite vertices lie on a parallel line, they can be shown to have equal area using the same principles or by translating one triangle.)
Corollary: Area of a Triangle Formula
This theorem leads directly to the standard formula for the area of a triangle.
Statement: The area of a triangle is equal to half the product of its base and the corresponding height.
Formula: $\text{Area of Triangle} = \frac{1}{2} \times \text{Base} \times \text{Height}$
Derivation of the Formula
Consider any triangle $\triangle ABC$ with base $\overline{BC}$. Let $h$ be the height corresponding to base $\overline{BC}$, which is the perpendicular distance from vertex A to the line containing $\overline{BC}$.
Construct a parallelogram DBCF on the same base $\overline{BC}$ and between the same parallels (the line containing $\overline{BC}$ and the line through A parallel to $\overline{BC}$). This can be done by drawing a line through B parallel to $\overline{AC}$ to meet the line through A parallel to $\overline{BC}$ at D. Quadrilateral ABDC is a parallelogram. $\triangle ABC$ and parallelogram ABDC are on the same base $\overline{BC}$ (or AB or AC depending on construction) and between the same parallels.
Let's consider $\triangle ABC$ with base $\overline{BC}$ and height $h$. Construct a parallelogram BCED on the base $\overline{BC}$ and between the parallels $\overline{BC}$ and the line through A. $D$ and $E$ will be on the line through A parallel to BC. This construction requires more careful definition of D and E.
A more standard way to derive the formula is to consider a rectangle with the same base and height.
Consider $\triangle ABC$ with base $\overline{BC}$ and height $h$ (perpendicular from A to BC). Construct a parallelogram XBCY on the base $\overline{BC}$ and between the parallel lines (BC and the line through A) such that its height is $h$. For instance, draw a line through A parallel to BC. Draw perpendiculars from B and C to this line, meeting it at X and Y. Then XBCY is a rectangle with base BC and height h. The area of this rectangle is $BC \times h$.
The relationship between $\triangle ABC$ and parallelogram XBCY (or a general parallelogram on the same base and height) is not as direct as using the fact that a parallelogram is divided into two equal area triangles by a diagonal.
Let's use the relationship shown in the previous theorem proof: a triangle is half the area of a parallelogram on the same base and between the same parallels.
Consider $\triangle ABC$ with base $\overline{BC}$ and height $h$. We can form a parallelogram on the base $\overline{BC}$ and height $h$. For example, draw a line through A parallel to $\overline{BC}$. From B, draw a line parallel to $\overline{AC}$ meeting the parallel line through A at D. Then ABDC is a parallelogram on base $\overline{BC}$ and between parallels $\overline{BC}$ and $\overline{AD}$. The height of this parallelogram corresponding to base $\overline{BC}$ is $h$.
By the theorem, $ar(\triangle ABC) = \frac{1}{2} ar(\text{parallelogram ABDC})$.
The area of parallelogram ABDC = Base $\times$ Height $= BC \times h$.
ar($\triangle ABC$) = $\frac{1}{2} (BC \times h)$
This proves that the area of a triangle is half the product of its base and the corresponding height.
Converse of the Theorem
The converse of this theorem is also true and is useful for proving that two lines are parallel.
Statement: Triangles on the same base (or equal bases) having equal areas lie between the same parallels.
More formally: If two triangles $\triangle ABC$ and $\triangle DBC$ are on the same base $\overline{BC}$ and have $ar(\triangle ABC) = ar(\triangle DBC)$, then the line segment $\overline{AD}$ is parallel to the line segment $\overline{BC}$.
Area Relationships for Triangles and Parallelograms
Based on the theorems concerning figures on the same base and between the same parallels, several important relationships regarding the areas of triangles and parallelograms can be established. These relationships are crucial for solving area-related problems in geometry.
1. Triangle and Parallelogram on Same Base and Between Same Parallels
This theorem connects the area of a triangle to the area of a parallelogram when they are positioned in a specific way relative to each other.
Theorem: If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram.
Given:
A parallelogram ABCD and a triangle $\triangle EBC$.
They share the same base $\overline{BC}$.
They lie between the same parallel lines, $\overline{BC}$ and a line through A, D, E (let's call it line AE, so $BC \parallel AE$).
To Prove:
$ar(\triangle EBC) = \frac{1}{2} ar(ABCD)$.
Construction:
Draw a line through C parallel to $\overline{BE}$ to meet the line AE at F. (This creates parallelogram EBCF).
Proof:
Consider the quadrilateral EBCF.
We are given that $\overline{EBC}$ is a triangle between parallels $\overline{BC}$ and $\overline{AE}$. So, $\overline{BC} \parallel \overline{EF}$ (since E and F lie on the line AE).
By construction, $\overline{CF} \parallel \overline{BE}$.
Since both pairs of opposite sides are parallel, EBCF is a parallelogram.
Now, consider parallelogram EBCF and its diagonal $\overline{EC}$. The diagonal divides the parallelogram into two congruent triangles, $\triangle EBC$ and $\triangle EFC$.
Thus, the area of $\triangle EBC$ is half the area of parallelogram EBCF.
ar($\triangle EBC$) = $\frac{1}{2}$ ar(parallelogram EBCF)
... (1)
Now, consider parallelogram ABCD and parallelogram EBCF.
Both parallelograms are on the same base $\overline{BC}$.
Both parallelograms lie between the same parallel lines, $\overline{BC}$ and the line containing AE and F.
By the theorem "Parallelograms on the same base and between the same parallels are equal in area" (Section I3):
ar(parallelogram ABCD) = ar(parallelogram EBCF)
... (2)
Substitute the equality from equation (2) into equation (1):
ar($\triangle EBC$) = $\frac{1}{2}$ ar(parallelogram ABCD)
Thus, the area of the triangle is half the area of the parallelogram when they share the same base and are between the same parallels.
2. Median of a Triangle and Area
This property demonstrates how a median affects the area distribution within a triangle.
Theorem: A median of a triangle divides it into two triangles of equal areas.
Given:
A triangle $\triangle ABC$.
$\overline{AD}$ is a median of $\triangle ABC$, where D is the mid-point of side $\overline{BC}$.
Thus, $BD = DC$.
To Prove:
The area of $\triangle ABD$ is equal to the area of $\triangle ADC$, i.e., $ar(\triangle ABD) = ar(\triangle ADC)$.
Construction:
Draw an altitude (perpendicular line segment) $\overline{AN}$ from vertex A to the side $\overline{BC}$. N lies on the line containing $\overline{BC}$.
Proof:
The area of any triangle is given by the formula: Area = $\frac{1}{2} \times \text{Base} \times \text{Height}$.
Consider $\triangle ABD$. Its base can be taken as $\overline{BD}$. The corresponding height is the perpendicular distance from vertex A to the line containing the base $\overline{BD}$. Since $\overline{BD}$ is part of the line $\overline{BC}$, the height is the length of the altitude $\overline{AN}$.
ar($\triangle ABD$) = $\frac{1}{2} \times BD \times AN$
... (1)
Now consider $\triangle ADC$. Its base can be taken as $\overline{DC}$. The corresponding height is the perpendicular distance from vertex A to the line containing the base $\overline{DC}$. Since $\overline{DC}$ is part of the line $\overline{BC}$, the height is the length of the altitude $\overline{AN}$.
ar($\triangle ADC$) = $\frac{1}{2} \times DC \times AN$
... (2)
We are given that $\overline{AD}$ is a median, which means D is the mid-point of $\overline{BC}$. Therefore, the lengths of the segments $\overline{BD}$ and $\overline{DC}$ are equal.
BD = DC
(Given, D is midpoint of BC)
Substitute $DC$ for $BD$ in equation (1):
ar($\triangle ABD$) = $\frac{1}{2} \times DC \times AN$
... (3)
Comparing equation (3) with equation (2), we see that the expressions for $ar(\triangle ABD)$ and $ar(\triangle ADC)$ are identical.
ar($\triangle ABD$) = ar($\triangle ADC$)
Thus, a median divides a triangle into two triangles of equal areas.
3. Converse Property for Triangles of Equal Areas
This is the converse of the theorem stated in Section I4.
Converse Property: Triangles on the same base (or on equal bases in the same line) having equal areas lie between the same parallels.
Formally: If two triangles, $\triangle ABC$ and $\triangle DBC$, are on the same base $\overline{BC}$ and $ar(\triangle ABC) = ar(\triangle DBC)$, then the line joining their vertices opposite the base (the line containing $\overline{AD}$) is parallel to the common base $\overline{BC}$.
4. Converse Property for Parallelograms of Equal Areas
This is the converse of the theorem stated in Section I3.
Converse Property: Parallelograms on the same base (or on equal bases in the same line) having equal areas lie between the same parallels.
Formally: If two parallelograms, ABCD and EBCF, are on the same base $\overline{BC}$ and $ar(ABCD) = ar(EBCF)$, then the line joining their vertices opposite the base (the line containing A, D, E, F) is parallel to the common base $\overline{BC}$.
Example 1. Show that a median of a triangle divides it into two triangles of equal areas.
Answer:
This was proved as Theorem 2 under this section. The proof is repeated here for clarity as requested by the input structure.
Given:
A triangle $\triangle ABC$ with median $\overline{AD}$, so D is the mid-point of $\overline{BC}$.
To Prove:
$ar(\triangle ABD) = ar(\triangle ADC)$.
Construction:
Draw an altitude $\overline{AN}$ from vertex A to the side $\overline{BC}$.
Proof:
Area of $\triangle ABD = \frac{1}{2} \times \text{base } \overline{BD} \times \text{height } \overline{AN}$
ar($\triangle ABD$) = $\frac{1}{2} \times BD \times AN$
Area of $\triangle ADC = \frac{1}{2} \times \text{base } \overline{DC} \times \text{height } \overline{AN}$
ar($\triangle ADC$) = $\frac{1}{2} \times DC \times AN$
Since $\overline{AD}$ is the median, D is the mid-point of $\overline{BC}$.
BD = DC
(Given, D is midpoint of BC)
Substituting $DC$ for $BD$ in the expression for $ar(\triangle ABD)$:
ar($\triangle ABD$) = $\frac{1}{2} \times DC \times AN$
Comparing this with the expression for $ar(\triangle ADC)$, we find they are equal:
ar($\triangle ABD$) = ar($\triangle ADC$)
Hence proved.
Example 2. ABCD is a parallelogram and P is any point on side CD. If $ar(\triangle DPA) = 15$ cm² and $ar(\triangle APC) = 20$ cm², find the area of $\triangle APB$.
Answer:
Given:
ABCD is a parallelogram.
P is a point on side $\overline{CD}$.
$ar(\triangle DPA) = 15 \text{ cm}^2$
$ar(\triangle APC) = 20 \text{ cm}^2$
To Find:
The area of $\triangle APB$, i.e., $ar(\triangle APB)$.
Solution:
Since P lies on the side $\overline{CD}$, the vertices D, P, and C are collinear. $\triangle DPA$ and $\triangle APC$ share a common vertex A, and their bases $\overline{DP}$ and $\overline{PC}$ are collinear along the line $\overline{CD}$. The sum of the areas of these two triangles is equal to the area of $\triangle ADC$.
Let $h$ be the perpendicular distance from A to the line $\overline{CD}$ (which is the height of the parallelogram corresponding to base CD). The area of $\triangle DPA = \frac{1}{2} \times DP \times h$ and the area of $\triangle APC = \frac{1}{2} \times PC \times h$.
ar($\triangle DPA$) + ar($\triangle APC$) = $\frac{1}{2} \times DP \times h + \frac{1}{2} \times PC \times h$
$= \frac{1}{2} \times (DP + PC) \times h$
Since P is on $\overline{CD}$, $DP + PC = DC$.
$= \frac{1}{2} \times DC \times h$
The area of $\triangle ADC$ is given by $\frac{1}{2} \times DC \times h$.
ar($\triangle DPA$) + ar($\triangle APC$) = ar($\triangle ADC$)
... (1)
We are given $ar(\triangle DPA) = 15$ and $ar(\triangle APC) = 20$. Substituting these values into equation (1):
ar($\triangle ADC$) = $15 + 20 = 35 \text{ cm}^2$
... (2)
Since ABCD is a parallelogram and $\overline{AC}$ is a diagonal, the diagonal divides the parallelogram into two triangles of equal area:
ar($\triangle ADC$) = $\frac{1}{2}$ ar(ABCD)
From equation (2), we have:
$\frac{1}{2}$ ar(ABCD) = $35 \text{ cm}^2$
... (3)
Now consider $\triangle APB$. This triangle and the parallelogram ABCD are on the same base $\overline{AB}$ and lie between the same parallel lines $\overline{AB}$ and $\overline{CD}$ (since P is on $\overline{CD}$).
By the theorem "If a triangle and a parallelogram are on the same base and between the same parallels, then the area of the triangle is half the area of the parallelogram" (Theorem 1 in this section):
ar($\triangle APB$) = $\frac{1}{2}$ ar(ABCD)
... (4)
Comparing equation (3) and equation (4), we see that both $ar(\triangle ADC)$ and $ar(\triangle APB)$ are equal to $\frac{1}{2}$ ar(ABCD).
Therefore:
ar($\triangle APB$) = ar($\triangle ADC$)
Substituting the value from equation (2):
ar($\triangle APB$) = $35 \text{ cm}^2$
The area of $\triangle APB$ is 35 cm².
Alternate Method:
As discussed earlier, for any point P on side CD of parallelogram ABCD, the following relationship holds:
ar($\triangle APB$) = ar($\triangle DPA$) + ar($\triangle BPC$)
From the areas of $\triangle DPA$ and $\triangle APC$ sharing vertex A and collinear bases on DC, we can find the ratio of their bases. Let $h$ be the altitude from A to DC.
ar($\triangle DPA$) = $\frac{1}{2} \times DP \times h = 15$
... (5)
ar($\triangle APC$) = $\frac{1}{2} \times PC \times h = 20$
... (6)
Dividing (5) by (6):
$\frac{ar(\triangle DPA)}{ar(\triangle APC)} = \frac{\frac{1}{2} \times DP \times h}{\frac{1}{2} \times PC \times h} = \frac{DP}{PC}$
$\frac{15}{20} = \frac{DP}{PC} \implies \frac{DP}{PC} = \frac{3}{4}$
Now, consider $\triangle BPC$. Its base is $\overline{PC}$ and its height is the perpendicular distance from B to the line $\overline{CD}$. Since ABCD is a parallelogram, the perpendicular distance from B to $\overline{CD}$ is the same as the perpendicular distance from A to $\overline{CD}$, which is $h$.
ar($\triangle BPC$) = $\frac{1}{2} \times PC \times h$
... (7)
Comparing (6) and (7), we see they have the same expression:
ar($\triangle BPC$) = ar($\triangle APC$) = $20 \text{ cm}^2$
Now, use the relationship $ar(\triangle APB) = ar(\triangle DPA) + ar(\triangle BPC)$.
ar($\triangle APB$) = $15 + 20 = 35 \text{ cm}^2$
Both methods give the same result.