Question

In the figure, C is the mid point of AB, $\angle BAD=\angle CBE,\angle ECA=\angle DCB$.
Prove that (a) $\Delta DAC\cong \Delta EBC$ (b) DA = EB.

Answer 1

(a) In given figure $C$ is the mid point of $AB$

$\angle BCE=\angle DCB+\angle ECD$ ..... $\left(i\right)$

$\angle DCA=\angle ECA+\angle ECD$ ...... $\left(ii\right)$

$\angle ECA=\angle DCB$ ..... (given )

$\therefore \angle BCE=\angle DCA$

Also, $\angle BAD=\angle CBE$ ....... (Given)

and $AC=BC$ ....... (Given)

$\therefore \Delta DAC\cong \Delta EBC$

(b) The $\Delta DAC\cong \Delta EBC$

Then all corresponding angles and sides are equal

Then $DA=EB$