# Example 3

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 3 ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD ∥ AB. Show that ∠ DAC = ∠ BCA and Given: ∆ ABC where AB = AC AD bisects ∠ PAC, & CD ∥ AB To prove: ∠ DAC = ∠ BCA Proof: AD bisects ∠ PAC Hence ∠ PAD =∠DAC = 1/2 ∠ PAC Also, given AB = AC ∴ ∠ BCA = ∠ ABC For ∆ ABC , ∠ PAC is an exterior angle So, ∠ PAC = ∠ ABC + ∠ BCA ∠ PAC = ∠ BCA + ∠ BCA ∠ PAC = 2∠ BCA 1/2∠ PAC = ∠ BCA ∠ BCA = 1/2∠ PAC ∠ BCA = ∠ DAC Hence proved Example 3 ABC is an isosceles triangle in which AB = AC. AD bisects exterior angle PAC and CD ∥ AB .Show that (ii) ABCD is a parallelogram. In previous part we proved , ∠ DAC = ∠ BCA For lines BC , AD with transversal AC ∠ DAC & ∠ BCA are alternate interior angles and they are equal So, BC ∥ AD Now, In ABCD BC ∥ AD & AB ∥ CD Since both pairs of opposite sides of quadrilateral ABCD are parallel. ABCD is a parallelogram.

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .