Question

Given :In Δ ABC: BD is perpendicular to AC, CE is perpendicular to AB and BD=CE.Prove:ΔABC is isosceles. refer to the pic

Answer 1

Consider triangles BEC and CDB
<E=<D
BD=EC  (given)
BC=BC  (common)
hence,triangle (BEC=CDB) by RHS congruence criteria
i.e.<B=<C
so, AC=AB, (sides opposite to equal angles are equal)
thus,it is proved that ABC is an isosceles triangle

Answer 2

In triangles BEC and BDC, 
BC is common
BD = CE (given)
<CEB = <BDC

Thus triangles BEC and BDC are congruent.
or, <ABC = <ACB (cpct)
Therefore, AB = AC (Opposite sides of equal angles are equal)
So, we can say that ABC is isosceles.

Hope that helps !!