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

Question

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

Rishabh Mittal · Accepted Answer

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

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

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