In Figure, bisectors of `/_B` and `/_D` of quadrilateral `A B C D` meet `C D` and `A B` produced at `Pa n d` Q respectively. Prove that : ` /_ P + /_Q= 1/2( /_ ABC + /_ ADC)`

Explanation

To prove that \( \angle P + \angle Q = \frac{1}{2}(\angle ABC + \angle ADC) \), we will follow these steps:

### Step 1: Analyze Triangle PBC
In triangle \( PBC \), we know that the sum of angles is \( 180^\circ \). Therefore, we can write:
\[ \angle CPB + \angle PBC + \angle BCP = 180^\circ \]
Let \( \angle CPB = \angle P \) and \( \angle PBC = \frac{1}{2} \angle ABC \) (since \( BP \) is the bisector of \( \angle ABC \)). Thus, we can rewrite the equation as:
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