Question

In $∆$CBS, seg BC $\cong$ seg SC ray CE bisects $\angle$DCS.
ray SE || ray BC. Prove that $\square$ CBSE is a parallelogram (see fig. 5.6)

Answer 1

We can see that $\angle$DCS = $\angle$CBS + $\angle$CSB (Exterior angle property)
$\Rightarrow$ $\angle$DCE + $\angle$ECS = $\angle$CBS + $\angle$CSB
$\Rightarrow$$\angle$CES + $\angle$ECS = $\angle$CBS + $\angle$CSB (CE bisects $\angle$DCE)
$\Rightarrow$ 2$\angle$ECS = $\angle$CSB + $\angle$CSB (Angles opposite to equal sides are equal )
$\Rightarrow$2$\angle$ECS = 2$\angle$CSB
$\Rightarrow$ $\angle$ECS = $\angle$CSB

Since, alternate angles are equal, CE and BS are parallel.
Since, each pair of opposite sides of a quadrilateral are parallel, $\square$CBSE is a parallelogram.