Question

In fig. BP ⊥ AC, CQ ⊥ AB, A–P–C and A–Q–B then show that ΔAPB and ΔAQC are similar.

In ΔAPB and ΔAQC

∠APB = `square`°   ...(i) 

∠AQC = `square`°   ...(ii)

∠APB ≅ ∠AQC   ...[From (i) and (ii)]

∠PAB ≅ ∠QAC   ...`square`

ΔAPB ~ ΔAQC   ...`square`

Answer 1

In ΔAPB and ΔAQC

∠APB = \[\boxed{90°}\]   ...(i)

∠AQC = \[\boxed{90°}\]   ...(ii)

∠APB ≅ ∠AQC   ...[From (i) and (ii)]

∠PAB ≅ ∠QAC   ...\[\boxed{[\text{Common angle}]}\]

ΔAPB ~ ΔAQC   ...\[\boxed{[\text{AA test of similarity}]}\]