Question

Please solve this question.
Subject: Mathematics
Grade: X
Topic: Triangles

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Answer 1

Given Question :-

In triangle ABC, DE is parallel to BC and BE bisects ∠ABC and CD bisects ∠ACB. Prove that triangle ABC is isosceles.

$\green{\large\underline{\sf{Given- }}}$

In triangle ABC,

• DE is parallel to BC.

• BE bisects ∠ABC

• CD bisects ∠ACB.

$\blue{\large\underline{\sf{To\:prove-}}}$

• Triangle ABC is isosceles

$\red{\large\underline{\sf{Solution-}}}$

Given that,

➢In triangle ABC,

• DE is parallel to BC.

We know,

Basic Proportionality Theorem,

If a line is drawn parallel to one side of a triangle, intersects the other two lines in distinct points, then the other two sides are divided in the same ratio.

So, By using Basic Proportionality Theorem,

$\red{\bf\implies \:\boxed{ \rm{ \: \frac{AD}{DB} = \frac{AE}{EC}}}} - - - (1)$

Also, given that

In triangle ABC

• BE bisects ∠ABC

We know,

Internal Angle bisector theorem

This theorem states that the bisector of an angle of a triangle divides the opposite side in the ratio of the sides containing the angle.

So, by using Internal Angle bisector theorem, we have

$\red{\bf\implies \:\boxed{ \rm{ \: \frac{AB}{BC} = \frac{AE}{EC}}}} - - - (2)$

Again, Given that,

In triangle ABC

• CD bisects ∠ACB

So, By using Internal Angle Bisector Theorem, we have

$\red{\bf\implies \:\boxed{ \rm{ \: \frac{AC}{BC} = \frac{AD}{DB}}}} - - - (3)$

On combining equation (1), (2) and (3), we get

$\rm :\longmapsto\:\dfrac{AB}{BC} = \dfrac{AC}{BC}$

$\bf\implies \:AB = AC$

$\bf\implies \: \triangle \: ABC \: is \: isosceles.$

Hence, Proved.

More to know

1. Pythagoras Theorem :-

This theorem states that : In a right-angled triangle, the square of the longest side is equal to sum of the squares of remaining sides.

2. Converse of Pythagoras Theorem :-

This theorem states that : If the square of the longest side is equal to sum of the squares of remaining two sides, angle opposite to longest side is right angle.

3. Area Ratio Theorem :-

This theorem states that :- The ratio of the area of two similar triangles is equal to the ratio of the squares of corresponding sides.