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.