Question

AB = AD = AC. Prove angle BCD = 90⁰​

Answer 1

Answer:

Given: In ∆ABC, AB = AC, side BA is produced to D such that AB = AD. To prove: ∠BCD = 90° Construction: Join CD. Proof: In ∆ABC given that AB = AC => ∠ACB = ∠ABC Now AB = AD AD = AC => ∠ACD = ∠ADC …(2) [Opposite angles of equal sides] Adding Eqn. (1) & Eqn. (2) ∠ACB + ∠ACD = ∠ABC + ∠ADC => ∠BCD = ∠ABC + ∠BDC [ v ∠ADC = ∠BDC] => ∠BCD + ∠BCD = ∠ABC + ∠BCD + ∠BDC => 2∠BCD = 180° [Adding ∠BCD on both sides] [By angle sum property of ∆] ∠BCD = 180/2 [∠ABC + ∠BCD + ∠BDC = 180°] ∠BCD = 90° So ∠BCD is a right angle.Read more on Sarthaks.com - https://www.sarthaks.com/857362/abc-is-isosceles-triangle-which-side-ba-is-produced-such-that-ad-show-that-bcd-right-angle?show=857370#a857370

Step-by-step explanation:

Given: In ∆ABC, AB = AC, side BA is produced to D such that AB = AD. To prove: ∠BCD = 90° Construction: Join CD. Proof: In ∆ABC given that AB = AC => ∠ACB = ∠ABC Now AB = AD AD = AC => ∠ACD = ∠ADC …(2) [Opposite angles of equal sides] Adding Eqn. (1) & Eqn. (2) ∠ACB + ∠ACD = ∠ABC + ∠ADC => ∠BCD = ∠ABC + ∠BDC [ v ∠ADC = ∠BDC] => ∠BCD + ∠BCD = ∠ABC + ∠BCD + ∠BDC => 2∠BCD = 180° [Adding ∠BCD on both sides] [By angle sum property of ∆] ∠BCD = 180/2 [∠ABC + ∠BCD + ∠BDC = 180°] ∠BCD = 90° So ∠BCD is a right angle.Read more on Sarthaks.com - https://www.sarthaks.com/857362/abc-is-isosceles-triangle-which-side-ba-is-produced-such-that-ad-show-that-bcd-right-angle?show=857370#a857370