Plz answer Share with your friends Share 1 Varun Rawat answered this Since, AD bisects ∠PAC, then∠PAD = ∠DAC .....1Now, in ∆ABC, ∠PAC = ∠ACB + ∠ABC Exterior angle theorem⇒2∠DAC = ∠ACB + ∠ABC .....2 using 1In ∆ABC, AB = AC⇒∠ACB = ∠ABC angles opposite to equal sides are equalNow, from 2, we get 2∠DAC = ∠ACB + ∠ACB⇒2∠DAC = 2∠ACB⇒∠DAC = ∠ACBBut ∠DAC and ∠ACB are alternate interior angles made by the lines AD and BC with transversal AC and are equal.So, AD∥BC. 1 View Full Answer Hisham Hafeez answered this Hello Answer Given:AC is the diagonal of Quadrilateral AB = DC therefore AB II DC To Prove : AD II BC Proof: In Triangle ABC and ADC AC = AC (Common) AB=DC (given) Angle BAC = Angle DCA (Alt Angles) Therefore Triangle ABC Congruent to Triangle ADC BC = AD (CPCT) Hope you understood this question Thanks 0 Hisham Hafeez answered this Sorry To prove : AD = BC 1 Advik answered this Given: AB=AC=DC To prove: AD||BC Proof: In triangle ABC, Angle ABC= angle ACB ( angles opp. to equal sides are equal) Angle PAC= angle ABC+ angle ACB ( exterior angle property) But angle PAC is also = angle PAD+ angle CAD Therefore, angle ABC + angle ACB = angle PAD + angle CAD 2(angle ACB)=2(angle CAD) Angle ACB= angle CAD Since 2 lines AD and BC are intersected by a transversal AC and a pair of alternate angles between them is equal therefore AD||BC Hence proved 0
