In triangle ABC - D is a point on AB and E is a point on AC such that DE BC- If AD - AB - AE - x- Then x isA- ACB- CDC- BAD- AE

The correct option is A ACIn ABC, DE ∥ BC nbsp; nbsp; nbsp; (Given) Therefore, AD/AB= AE/AC nbsp; nbsp; [By Basic Proportionality Theorem] Compari ...

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Standard X

Mathematics

Thales Theorem

In triangle A...

Question

In triangle $A B C, D$ is a point on $A B$ and $E$ is a point on $A C$ such that $D E | | B C$ . If $\frac{A D}{A B} = \frac{A E}{x}$ , Then $x$ is _______.

A C

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B A

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A E

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C D

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Solution

The correct option is A $A C$
In $△$ ABC,

DE $∥$ BC (Given)

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

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

$⟹ x = A C$

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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 _______.

The correct option is A ACIn △ABC, DE ∥ BC (Given) Therefore, ADAB=AEAC [By Basic Proportionality Theorem] Comparing the above with ADAB = AEx ⟹x=AC

In triangle $A B C, D$ is a point on $A B$ and $E$ is a point on $A C$ such that $D E | | B C$ . If $\frac{A D}{A B} = \frac{A E}{x}$ , Then $x$ is _______.

The correct option is A $A C$
In $△$ ABC,

DE $∥$ BC (Given)

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

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

$⟹ x = A C$