Question

In triangle ABC angle B=angle C,D and E are the point on AB and AC such that BD= CE prove that DE parallel BC

Answer 1

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In Triangle ABC we have :-

∠B = ∠C

AC = AB     ( sides opposite to equal angles are equal )

AE + EC = AD + DB

AE + CE = AD + BD

AE + CE = AD + CE ( BD = CE)

AE = AD

Thus we are having :-

AD = AE and BD = CE

Therefore, AD/BD = AE/CE

AD/DB = AE/EC

DE // BC ( By converse of Thales Theorem )

$\huge{\boxed{\sf{HENCE\:PROVED!}}}$

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Answer 2

Given :

In triangle ABC angle B=angle C,D and E are the point on AB and AC such that BD= CE

To Find :

prove that DE parallel B

Solution:

In ΔABC

$\angle B = \angle C$

Opposite sides of equal angles are equal

So, AB = AC

Now we are given that D and E are the point on AB and AC such that BD= CE

AB = AC

AD+BD=AE+CE

So, AD+CE= AE+CE

AD=AE

So, D and E are the midpoints of AB and AC

So, By mid point theorem(The Midpoint Theorem states that the line segment joining the midpoints of any two sides of a triangle is parallel to the third side and equal to half of the third side)

So, DE||BC

Hence Proved