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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

Where is the pic????


Consider triangles BEC and CDB
BD=EC  (given)
BC=BC  (common)
hence,triangle (BEC=CDB) by RHS congruence criteria
so, AC=AB, (sides opposite to equal angles are equal)
thus,it is proved that ABC is an isosceles triangle
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In triangles BEC and BDC, 
BC is common
BD = CE (given)

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 !!
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