Q2 of 49 Page 58

Suppose ABC is a triangle and D is the midpoint of BC. Assume that the perpendiculars from D to AB and AC are of equal length. Prove that ABC is isosceles.

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In Δ ABC, D is the midpoint on BC


Let PD and QD be the two perpendiculars drawn from D on AB and AC respectively


In Δ BPD and Δ CQD we have


PD = QD(Given)


∠DPB = ∠DQC = 900(Perpendiculars)


BD = DC(D is a midpoint)


So Δ BPD and Δ CQD are congruent by R.H.S. axiom of congruency


So according to Corresponding Parts of Congruent triangles we get


∠PBD = ∠QCD


Since the two angles of ABC are equal so Δ ABC isosceles.


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