Name: Areas of Similar Triangles
Uploaded: 2019-07-19
Duration: 6 min 54 s
Description: The corresponding sides of similar triangles are proportional. Are the ratios of areas of similar triangles the same as that of their sides? Watch this video...

Question

State and prove Basic Proportionality Theorem.

Accepted Answer

Basic Proportionality theorem: If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides of the triangle in proportion.

Let $A B C$ be the triangle and the line $l$ parallel to $B C$ intersect $A B$ at $D$ and $A C$ at $E$ .

To prove: $\frac{A D}{D B} = \frac{A E}{E C}$

Construction: Join $B E$ and $C D$ . Draw $E F ⊥ A B, D G ⊥ A C$ .

Proof: $E F$ is the height of $△ A D E$ and $△ D B E$ since $E F ⊥ A B$ .

Area of a triangle $= \frac{1}{2} \times Base \times Height$

$∴ \frac{Area of △ A D E}{Area of △ D B E} = \frac{\frac{1}{2} \times E F \times A D}{\frac{1}{2} \times E F \times D B} = \frac{A D}{B D}$ _____ $(1)$

We know that triangles on same base and between the same parallel straight line are equal in area.

But $△ D B E$ and $△ D C E$ are the same base $D E$ and between the same parallel straight line $B C$ and $D E$ .

$∴ a r (△ D B E) = a r (△ D C E)$

Similarly, $D G$ is the height of $△ A D E$ and $△ D C E$ since $D G ⊥ A C$ .

$∴ \frac{Area of △ A D E}{Area of △ D C E} = \frac{\frac{1}{2} \times D G \times A E}{\frac{1}{2} \times D G \times E C}$

$\Rightarrow \frac{Area of △ A D E}{Area of △ D B E} = \frac{A E}{E C}$ _____ $(2)$

From $(1)$ and $(2)$ , we get,

$\frac{A D}{B D} = \frac{A E}{E C}$

Hence, proved.

State and prove Basic Proportionality Theorem.

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