### In the figure, it is given that $AC = BC, \angle 4 = 2 \space \angle 1$ and $\angle 3 = 2 \space \angle 2$. Prove that $\triangle ADC \cong \triangle BEC.$ C A B E D 1 2 4 3

1. We are given that $AC = BC, \angle 4 = 2 \space \angle 1$ and $\angle 3 = 2 \space \angle 2.$
2. We need to find a triangle congruent to $\triangle ADC.$
3. In $\triangle ABC$, we have \begin{aligned} &AC = BC &&\text{[Given]} \\ \implies &\angle CAB = \angle CBA && \text{[Angles opposite to equal sides are equal]} &&\ldots (1) \end{aligned} Also, \begin{aligned} &\angle 4 = \angle 3 && \text{[Vertically opposite angles]} \\ \implies &2 \space \angle 1 = 2 \space \angle 2 && \text{[As } \angle 4 = 2 \space \angle 1 \text{ and } \angle 3 = 2 \space \angle 2] \\ \implies &\angle 1 = \angle 2 && \ldots (2) \end{aligned} Subtracting equation (2) from equation (1), we have \begin{aligned} &\angle CAB - \angle 1 = \angle CBA - \angle 2 \\ \implies &\angle CAD = \angle CBE && \ldots (3) \end{aligned}
4. In $\triangle ADC$ and $\triangle BEC$, we have \begin{aligned} &\angle ACB = \angle ACB &&\text{[Common]} \\ &AC = BC &&\text{[Given]} \\ &\angle CAD = \angle CBE &&\text{[From equation (3)]} \\ \therefore {\space} &\triangle ACD \cong \triangle BCE &&\text{[By ASA criterion]} \end{aligned}
5. Thus, $\bf { \triangle ACD \cong \triangle BCE}$.