Question

In the given figure, C is the mid-point of AB. If ∠DCA = ∠ECB and ∠DBC = ∠EAC, prove that DC = EC.
Figure

Answer 1

$$\begin{gathered} \mathrm{In}∆ECA\mathrm{and}∆DCB, \\ \angle EAC=\angle DBC\left(\mathrm{given}\right) \\ AC=BC\left(C\mathrm{is}\mathrm{the}\mathrm{mid}\mathrm{point}\mathrm{of}AB\right) \\ \mathrm{Now}, \\ \angle DCA\hspace{0.17em}=\angle ECB \\ \Rightarrow \angle DCA+\angle DCE=\angle ECB+\angle DCE \\ \Rightarrow \angle ACE\hspace{0.17em}=\angle BCD \\ \mathrm{Therefore},\mathrm{by}ASA\mathrm{congruency}\mathrm{criteria},∆ECA\cong ∆DCB. \\ \Rightarrow DC=EC\left(\text{C}\mathrm{orresponding}\mathrm{si}\text{d}\mathrm{es}\mathrm{of}\mathrm{congruent}\mathrm{triangles}\mathrm{are}\mathrm{equal}\right) \end{gathered}$$