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ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is isosceles.

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Given: D is the midpoint of BC and PD = DQ in a triangle ABC. 

To prove: ABC is isosceles triangle.

In △BDP and △CDQ 

PD = QD (Given) 

BD = DC (D is mid-point) 

∠BPD = ∠CQD = 90o 

By RHS Criterion: △BDP ≅ △CDQ 

BP = CQ … (i) (By CPCT) 

In △APD and △AQD 

PD = QD (given) 

AD = AD (common) 

APD = AQD = 90° 

By RHS Criterion: △APD ≅ △AQD 

So, PA = QA … (ii) (By CPCT) 

Adding (i) and (ii) 

BP + PA = CQ + QA 

AB = AC 

Two sides of the triangle are equal, so ABC is an isosceles.

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