Name: Proof of Basic Proportionality theorem
Uploaded: 2019-07-19
Duration: 3 min 10 s
Description: When a line is drawn parallel to one side, it intersects the other two sides of a triangle in the same ratio. Watch this video to learn more about them.

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 $D E$ .

$∴ △ D B E = △ D C E$

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

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

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

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

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

Hence, proved.

State and prove basic proportionality theorem.

Important Questions on Triangles

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