In ∆ABC, D and E are points on sides AB and AC respectively
Question:

In ∆ABC, D and E are points on sides AB and AC respectively such that AD ✕ EC = AE ✕ DB. Prove that DE || BC.

Solution:

Given: In $\triangle A B C, D$ and $E$ are points on sides $A B$ and $A C$ such that $A D \times E C=A E \times D B$

To Prove: $D E \| B C$

Proof:

Since $A D \times E C=A E \times D B$

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

$\Rightarrow \quad \frac{D B}{A D}+1=\frac{E C}{A E}+1$

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

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

$\therefore D E \| B C$ (Converse of basic proportionality theorem)

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