In the figure, it is given that ^@AC = BC, \angle 4 = 2 \space \ | Math Question and Answer

Answer:

Step by Step Explanation:

We are given that $A C = B C, ∠ 4 = 2 ∠ 1$ and $∠ 3 = 2 ∠ 2.$
We need to find a triangle congruent to $\triangle ADC.$
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 $\begin{aligned} &\angle CAB - \angle 1 = \angle CBA - \angle 2 \\ \implies &\angle CAD = \angle CBE && \ldots (3) \end{aligned}$
In $\triangle ADC$ and $\triangle BEC$ , we have $\begin{aligned} &\angle ACB = \angle ACB &&\text{[Common]} \\ &AC = BC &&\text{[Given]} \\ &\angle CAD = \angle CBE &&\text{[From equation (3)]} \\ \therefore {\space} &\triangle ACD \cong \triangle BCE &&\text{[By ASA criterion]} \end{aligned}$
Thus, $\bf { \triangle ACD \cong \triangle BCE}$ .

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