Inthe given figure , if AB =AC , D is a point on AC and E on AB such that AD = ED = EC = BC. Prove that (i) <A : <B= 1:3 (ii) <AED=<BCE Share with your friends Share 42 Rishabh Mittal answered this 1 ∠A=∠AED Since AED is isoceles triangle∠B=∠BEC Since BEC is isoceles triangle∠BEC+∠CED+∠AED=π Linear Pair∠A+∠B+∠CED=π ...1In triangle CED,∠CED+2∠EDC=π Since CED is isoceles triangle∠CED=π-2∠EDCSubstituting the value of ∠CED in 1 , we get∠A+∠B+π-2∠EDC=π∠A+∠B-2∠EDC=0 ...2∠EDC=∠DEA+∠A Exterior angle∠EDC=2∠A Since DEA is isosceles triangleSubstituting the value of ∠EDC in 2 , we get∠A+∠B-22∠A=03∠A=∠B∠A∠B=13Hence Proved.2 ∠ECA=∠EDC Since CED is isosceles triangle∠BCA=∠B Since ABC is isosceles triangle∠EDC = 2∠A Since EDA is isosceles triangleSo, ∠ECA=2∠A∠BCE=∠BCA-∠ECA∠BCE=∠B-2∠A∠BCE=3∠A-2∠A∠BCE=∠A ...3∠AED =∠A ...4 Since AEC is isosceles triangleFrom 3 and 4 , we get∠AED=∠BCE Hence Proved. 27 View Full Answer