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(we have drawn an triangle ABC and AD is the altitude)
Given : A triangle ABC  in which BD = DC

to prove : AB = AC or ABC is an isosceles triangle .
In triangle ABD and triangle ADC

AD = AD ( common)
∠ADB = ∠ADC (90° each)
BD = DC (given)

so Δ ABD ≡ ΔACD( by RHS)

So AB = AC (by cpct)
∴ Δ ABC is an isosceles triangle.
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First we take 2 triangles ADB & ACD..
                        angleADB=angleADC  (both 90)
                                     BD=CD        (given)
                         ADB = ACD       (ASA)
                        AB=AC                (C.P.C.T)
Hence ABC is isosceles triangle..

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