In an isosceles triangle \(ABC\), \(AB = AC\). \(AD\) bisects exterior angle \(PAC\) and \(CD\) parallel to \(AB\) (see Fig.).
Prove that
(i) \(∠ DAC = ∠ BCA \) and
(ii) The quadrilateral \(ABCD\) is a parallelogram.
\(∆ ABC\) is isosceles in which \(AB = AC\)
So, \(∠ ABC = ∠ \) (Angles opposite to equal sides)
Also, \(∠ PAC = ∠ ABC + ∠ ACB\) (Exterior angle of a triangle)
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, \(∠ DAC = ∠ ACB\)
(ii) To Prove: The quadrilateral \(ABCD\) is a parallelogram.
Proof:
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)
Now, both pairs of opposite sides of quadrilateral \(ABCD\) are .
So, The quadrilateral \(ABCD\) is a parallelogram.