bisector of angle b and angle d of a quadrilateral abcd meet cd and ab produced at point p and - Maths - Lines and Angles

Question

bisector of angle b and angle d of a quadrilateral abcd meet cd
and ab produced at point p and q respectively prove( angle p +angle
q=1/2 angle abc +angle adc)

Pronay · Accepted Answer

In Î”AQD , by angle sum property $∠ BAD+ ∠ Q+ \frac{1}{2} ∠ ADC= 180^{\circ} (1)$ In Î”CPB, by angle sum property, $∠ BCD+ ∠ P+ \frac{1}{2} ∠ ABC= 180^{\circ} (2)$ $(1) + (2) \Rightarrow$ $∠ BAD+ ∠ Q+ \frac{1}{2} ∠ ADC+ ∠ BCD+ ∠ P+ \frac{1}{2} ∠ ABC= 360^{\circ} \Rightarrow ∠ BAD+ ∠ Q+ \frac{1}{2} ∠ ADC+ ∠ BCD+ ∠ P+ \frac{1}{2} ∠ ABC= ∠ B AD+ ∠ ABC+ ∠ ADC+ ∠ BCD \Rightarrow ∠ P+ ∠ Q= \frac{1}{2} (∠ ABC+ ∠ ADC)$

bisector of angle b and angle d of a quadrilateral abcd meet cd and ab produced at point p and q respectively prove( angle p +angle q=1/2 angle abc +angle adc)