Question

From the following figure, Prove that:
$$AD=CE$$

Answer 1

In $$\triangle ACB,$$
$$AC=AC$$ [Given]
$$\therefore \angle ABC=\angle ACB$$......(i) [angles opp. to equal sides are equal]
$$\angle ACD=\angle ACB=180^{o}$$......(ii) [$$DCB$$ is a straight line ]
$$\angle ABC=\angle CBE=180^{o}$$......(iii) [$$ABE$$ is a straight line ]
Equating (ii) and (iii)
$$\angle ACD+\angle ACB=\angle ABC+\angle CBE$$
$$\angle ACD+\angle ACB=\angle ACB+\angle CBE$$ [From (i)]
$$\Rightarrow \angle ACD=\angle CBE$$

In $$\triangle ACD$$ and $$\triangle CBE,$$

$$DC=CB$$ [Given]
$$AC=BE$$ [Given]
$$\angle ACD=\angle CBE$$ [Proved Earlier]
$$\therefore \triangle ACD\cong \triangle CBE$$ [SAS criterian]
$$\Rightarrow AD=CE$$ [cpct]