Q3 of 59 Page 197

AD is median of the triangle ABC. The smallest distance of AD from B and C are BP and CQ respectively. Let’s proved that BP=CQ.

Let us first draw a diagram for the above given question.


In Δ BPD and Δ CQD


∠ BPD = ∠ CQD ………….. [As both the angles are 90°]


∠ BDP = ∠ CDQ ………….. [Vertically opposite angles]


BP = CD ………………………. [As AD is median to BC so BC is divided into two equal halves]


Hence Δ BPD is congruent to Δ CQD by Angle Angle Side test.


Since both the triangles are congruent, side BP = side CQ [By congruent parts of congruent triangle are equal]


Hence BP = CQ.


Hence Proved.


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