If triangle \(ABC\) is isosceles with \(AB = AC\), and \(AD\) bisects the exterior angle \(PAC\) while \(CD\) is parallel to \(AB\), prove that:
(i) \(\angle DAC = \angle BCA\), and
(ii) the quadrilateral \(ABCD\) is a parallelogram.
Proof:
(i) \(∆ ABC\) is isosceles in which \(AB = AC\) (Given)
So, \(∠ ABC = ∠ ACB\) []
Also, \(∠ PAC = ∠ ABC + ∠ ACB\)[]
That is, \(∠ PAC = 2∠ ACB\)---- (1)
Now, \(AD\) bisects \(∠ PAC\).
So, \(∠ PAC = 2∠ DAC\)----- (2)
Therefore, \(2∠ DAC = 2∠ ACB\) [From (1) and (2)]
That is, \(∠ DAC = ∠ ACB\)
Now, \(AD\) bisects \(∠ PAC\).
So, \(∠ PAC = 2∠ DAC\)----- (2)
Therefore, \(2∠ DAC = 2∠ ACB\) [From (1) and (2)]
That is, \(∠ DAC = ∠ ACB\)
(ii) Now, these equal angles form a pair of alternate angles when line segments \(BC\) and \(AD\) are intersected by a transversal \(AC\).
So, \(BC || AD\)
Also, \(BA || CD\) (Given)
So, \(BC || AD\)
Also, \(BA || CD\) (Given)
Now, both pairs of opposite sides of quadrilateral \(ABCD\) are .
So, \(ABCD\) is a parallelogram.