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 = [   ]°     ......(i)

∠AQC = [   ]°  ......(ii)

∠APB ≅ ∠AQC    .....[From (i) and (ii)]

∠PAB ≅ ∠QAC    .....[______]

ΔAPB ~ ΔAQC     .....[______]

Answer 1

In ΔAPB and ΔAQC

∠APB = 90°     ......(i)

∠AQC = 90°  ......(ii)

∠APB ≅ ∠AQC    .....[From (i) and (ii)]

∠PAB ≅ ∠QAC    .....[Common angle]

ΔAPB ~ ΔAQC     .....[AA test of similarity]