Question

In the given figure bp is perpendicular to ac,cq is perpendicular to ab, a-p-c a-q-b then prove that triangle apb is similar to triangle aqc

Answer 1

Δapb ≈ Δaqc

Step-by-step explanation:

bp is perpendicular to ac

=> ∠apb = 90°

cq is perpendicular to ab

=> ∠aqc = 90

now comparing

Δapb & Δaqc

∠apb  = ∠aqc  = 90°

∠a = ∠a   ( common angle)

=> Δapb ≈ Δaqc

QED

Proved

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