in Parallelogram ABCD bisector of adjacent angels A and B intersect each other at P prove that APB 90
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AD is parallel to BC and transversal AB intersects them.
∴∠a+∠b=180° (ADJACENT ANGLE SUM IS 180°)
1/2(∠a+∠b)=1/2×180°
∠PAB+∠PBA=90°
In ΔPAB
∠PAB+∠PBA+∠APB=180° (ANGLE SUM PROPERTY OF TRIANGLE)
∠APB=180°-90°
∠APB=90°
∴∠a+∠b=180° (ADJACENT ANGLE SUM IS 180°)
1/2(∠a+∠b)=1/2×180°
∠PAB+∠PBA=90°
In ΔPAB
∠PAB+∠PBA+∠APB=180° (ANGLE SUM PROPERTY OF TRIANGLE)
∠APB=180°-90°
∠APB=90°
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