Question

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

Answer 1

Given: C is the midpoint of AB.
Or AC = CB

$$\begin{gathered} \mathrm{Also},\angle DCA=\angle ECB\left(\mathrm{Given}\right) \\ \Rightarrow \angle DCA+\angle DCE=\angle ECB+\angle DCE \\ \Rightarrow \angle ACE=\angle BCD...\left(i\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{in}△\mathrm{DBC}\mathrm{and}△\mathrm{EAC},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{DBC}=\angle \mathrm{EAC}\left(\mathrm{Given}\right) \\ \mathrm{BC}=\mathrm{AC}\left(\mathrm{Given}\right) \\ \angle \mathrm{BCD}=\angle \mathrm{ACE}[\mathrm{From}(\mathrm{i}\left)\right] \\ \mathrm{i}.\mathrm{e}.,△\mathrm{DBC}\cong △\mathrm{EAC}(\mathrm{ASA}\mathrm{criterion}) \end{gathered}$$
∴ DC = EC (CPCT)
Hence, proved.