In -ABC- AD-DB-AE-EC and -ADE - -ACB- Then -ABC is an-

From Converse of Basic Proportionality theorem, Since AD/DB=AE/EC DE∥ BC. Given ∠ ADE = ∠ ACB — (1) Also ∠ ADE = ∠ ABC nbsp; nbsp;—–(2) (Transverse angles ...

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Standard X

Mathematics

Converse of Basic Proportionality Theorem

In ΔABC, AD/D...

Question

In $Δ A B C$ , $\frac{A D}{D B} = \frac{A E}{E C}$ and $∠ A D E$ = $∠ A C B$ . Then $Δ A B C$ is an.

equilateral triangle

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isosceles triangle

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scalene triangle

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Solution

The correct option is B isosceles triangle
From Converse of Basic Proportionality theorem,
Since $\frac{A D}{D B} = \frac{A E}{E C} ⟹ D E ∥ B C$ .
Given $∠$ ADE = $∠$ ACB ---- (1)
Also $∠$ ADE = $∠$ ABC -----(2)
(Transverse angles in parallel lines are equal)
From (1) & (2),
$\Rightarrow ∠ A B C = ∠ A C B$
Hence it is an "isosceles triangle".

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Similar questions

Consider $Δ$ ABC. If $\frac{A D}{D B} = \frac{A E}{E C}$ and $∠$ ADE = $∠$ ACB. Then $Δ$ ABC is an equilateral triangle.

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In ΔABC, ADDB=AEEC and ∠ADE = ∠ACB. Then ΔABC is an.

The correct option is B isosceles triangleFrom Converse of Basic Proportionality theorem, Since ADDB=AEEC⟹DE∥BC. Given ∠ ADE = ∠ ACB ---- (1) Also ∠ ADE = ∠ ABC -----(2) (Transverse angles in parallel lines are equal) From (1) & (2), ⇒∠ABC=∠ACB Hence it is an "isosceles triangle".

In $Δ A B C$ , $\frac{A D}{D B} = \frac{A E}{E C}$ and $∠ A D E$ = $∠ A C B$ . Then $Δ A B C$ is an.