In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If $\frac{A D}{A B}$ = $\frac{A E}{x}$ , Then $x$ is
___.

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Solution

In $△$ ABC,

DE $∥$ BC (Given)

Therefore, $\frac{A D}{A B}$ = $\frac{A E}{A C}$ [By Basic Proportionality Theorem]

Comparing the above with $\frac{A D}{A B}$ = $\frac{A E}{x}$

$x$ = AC

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In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If ADAB = AEx, Then x is ___.

In △ABC, DE ∥ BC (Given) Therefore, ADAB = AEAC [By Basic Proportionality Theorem] Comparing the above with ADAB = AEx x = AC

In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If $\frac{A D}{A B}$ = $\frac{A E}{x}$ , Then $x$ is
___.

In $△$ ABC,

DE $∥$ BC (Given)

Therefore, $\frac{A D}{A B}$ = $\frac{A E}{A C}$ [By Basic Proportionality Theorem]

Comparing the above with $\frac{A D}{A B}$ = $\frac{A E}{x}$

$x$ = AC