In the adjoining figure, AB = AC: CP||BA and AP is the bisector of `angle CAD. ` Prove that
(i) `angle PAC = angle BCA and `
(ii) ABCP is a parallelogram.

Explanation

In `triangle ABC, AB = AC rArr angle 1 = angle 2.`
Ext. `angle CAD = angle 1 + angle 2`
`rArr angle CAD =2(angle 2) " " [ because angle 1 = angle 2]`
`rArr 2 angle PAC = 2(angle 2)rArr angle PAC = angle 2 rArr angle PAC = angle BCA.`
Now , `angle PAC = angle BCA rArr AP||BC " " (because angle PAC and angle BCA " are alt. interior " angle)`
Also, CP||BA (given).
`therefore ` ABCP is a parallelogram.