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Angles in Same Segment of a Circle

In the figure...

Question

In the figure, given alongside, CP bisects angle ACB.

Show that DP bisects angle ADB.

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Solution

Given $CP$ bisects $\angle ACB$

$∴$ $\angle ACP = \angle PCB$

$\angle ACP = \angle ADP$ (angles in the same segment)

$\angle PCB = \angle PDB$ (angles in the same segment)

Therefore $\angle ADP = \angle PDB$

Hence $DP \ bisects \ \angle ADB$

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