Question

In an isosceles `Delta ABC`, the base AB is produced both wasys in P and Q such that `APxxBQ=AC^(2)`
Prove that `Delta ACP~ Delta BCQ`.

Answer 1

`CA=CB rArr angle CAB= angle CBA`
`rArr 180^(@)-angleCAB=180^(@)-angle CBA`
`rArr angle CAP= angle CBQ`.
Now, `APxxBQ=AC^(2) rArr (AP)/(AC)=(AC)/(BQ)`
`rArr (AP)/(AC)=(BC)/(BQ) [ :. AC=BC]`

Thus, `angle CAP= angle CBQ and (AP)/(AC)=(BC)/(BQ)`
`:. Delta ACP~Delta BCQ`.