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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.^@
Answer:
- We are given that ^@ AC = BC,
\angle 4 = 2 \space \angle 1 ^@ and
^@ \angle 3 = 2 \space \angle 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} ^@.
