Question

triangle ABC, line DE is drawn parallel to side BC such that AD/DB = AE/EC. Show that BAC is an isosceles

Answer 1

Given: A triangle ABC, line DE is drawn parallel to side BC such that

$\frac{AD}{DB}=\frac{AE}{EC}$.

One more condition should be there that is $\angle ADE=\angle ACB$.

To prove: Triangle ABC is isosceles triangle.

Proof: Since, line DE is parallel to BC.

By Basic Proportionality theorem,

$\frac{AD}{DB}=\frac{AE}{EC}$

Now, since line DE is parallel to BC.

$\angle ADE=\angle ABC$ (Corresponding angles)  (Equation 1)

Since it is given that $\angle ADE=\angle ACB$        (Equation 2)

Therefore, by equations 1 and 2,

$\angle ABC=\angle ACB$

Therefore, AB = AC

(Because sides opposite to the equal opposite angles are equal).

Hence, triangle ABC is an isosceles triangle.

Answer 2

Answer:

Given: A triangle ABC, line DE is drawn parallel to side BC such that

\frac{AD}{DB}=\frac{AE}{EC}

DB

AD

=

EC

AE

.

One more condition should be there that is \angle ADE=\angle ACB∠ADE=∠ACB .

To prove: Triangle ABC is isosceles triangle.

Proof: Since, line DE is parallel to BC.

By Basic Proportionality theorem,

\frac{AD}{DB}=\frac{AE}{EC}

DB

AD

=

EC

AE

Now, since line DE is parallel to BC.

\angle ADE=\angle ABC∠ADE=∠ABC (Corresponding angles) (Equation 1)

Since it is given that \angle ADE=\angle ACB∠ADE=∠ACB (Equation 2)

Therefore, by equations 1 and 2,

\angle ABC=\angle ACB∠ABC=∠ACB

Therefore, AB = AC

(Because sides opposite to the equal opposite angles are equal).

Hence, triangle ABC is an isosceles triangle.