Question

AM is the median of triangle abc, BP and CQ are perpendicular to AM and AM produced. Prove that BP=CQ

Answer 1

Answer:

Given: BP and CQ are medians of AB and AC respectively of triangle ABC

BP and CQ are produced to M and N such that BP = PM and CQ = QN

In △APM and △BPC,

AP=PC

PM=BP

∠APM=∠BPC ...(Vertically opposite angles)

therefore, △APM≅△BPC ...(SAS rule)

∠AMP=∠PBC ...(By cpct)

Similarly, △AQN≅△BPC

hence, ∠ANQ=∠QBC ..(By cpct)

Hence, N, A, M lie on a straight line.

NM=NA+AM=BC+BC=2BC

hence, A is the mid point of MN