In the adjoining figure `BP_|_AC, CQ_|_AB, A-P-C,A-Q-B`, then prove that `DeltaAPB` and `DeltaAQC` are similar.

Explanation

In `DeltaAPB` and `DeltaAQC, /_APB=square`………..I
`/_AQC=square`……..II
`:./_APB~=/_AQC`……………… From I and II
`/_PAB!=/_QAC`……………. (`square`)
`:.DeltaAPB~DeltaAQC`……………(AA test)
Ans:
In `DeltaAPB` and `DeltaAQC`
`/_APB=90^(@)` ......I
`/_AC=90^(@)`..........II
`:./_APB~=/_AQC`...........[From I and II]
`/_PAB~=/_QAC`... (Common angle)
`:.DeltaAPB~DeltaAQC`...............(AA test)