In -ABC- AD-DB-AE-EC and -ADE - -ACB-Prove that -ABC is an isosceles triangle- - -2 Marks-

By the Converse of Basic Proportionality theorem, Since AD/DB=AE/EC DE∥ BC [1 Mark] Given ∠ ADE = ∠ ACB ...(1) Also, ∠ ADE = ∠ ABC...(2) [As DE || ...

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Standard X

Mathematics

Basic Proportionality Theorem

In ΔABC, AD/D...

Question

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

Prove that $Δ A B C$ is an isosceles triangle. . [2 Marks]

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Solution

By the Converse of Basic Proportionality theorem,
Since $\frac{A D}{D B} = \frac{A E}{E C} ⟹ D E ∥ B C [1 M a r k]$
Given $∠ A D E = ∠ A C B . . . (1)$
Also, $∠ A D E = ∠ A B C . . . (2)$ [As DE || BC and corresponding angles are equal]

[0.5 Marks]
From (1) & (2),
$\Rightarrow ∠ A B C = ∠ A C B$
Hence it is an "isosceles triangle". [0.5 Marks]

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Similar questions

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

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 a/an

.

[ 3 Marks ]

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.

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

Q. In

Δ

ABC,

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

and

∠

ADE =

∠

ACB. Then

Δ

ABC is triangle.

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In ΔABC,ADDB=AEEC and ∠ADE=∠ACB. Prove that ΔABC is an isosceles triangle. . [2 Marks]

By the Converse of Basic Proportionality theorem, Since ADDB=AEEC⟹DE∥BC [1Mark] Given ∠ADE=∠ACB...(1) Also, ∠ADE=∠ABC...(2) [As DE || BC and corresponding angles are equal] [0.5 Marks] From (1) & (2), ⇒∠ABC=∠ACB Hence it is an "isosceles triangle". [0.5 Marks]

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

Prove that $Δ A B C$ is an isosceles triangle. . [2 Marks]