Question

$\text{In the following diagram, the bisectors of interior angles of the parallelogram PQRS en-close a quadrilateral ABCD.}$

$\text{Show that:}\left(i\right)\angle PSB+\angle SPB={90}^{\circ }\left(ii\right)\angle PBS={90}^{\circ }\left(iii\right)\angle ABC={90}^{\circ }\left(iv\right)\angle ADC={90}^{\circ }\left(v\right)\angle A={90}^{\circ }\left(vi\right)\text{ABCD is a rectangle}\text{Thus, the bisectors of the angles of a parallelogram enclose a rectangle.}$

Answer 1

$\text{Given : In parallelogram ABCD bisector of}\text{angles P and Q, meet at A, bisectors of}angleR.\text{ABCD as shown in the figure.}\text{To Prove :}\left(i\right)\angle PSB+\angle SPB={90}^{\circ }\left(ii\right)\angle PSB={90}^{\circ }\left(iii\right)\angle ABC={90}^{\circ }\left(iv\right)\angle ADC={90}^{\circ }\left(v\right)\angle A=9^{\circ }\left(vi\right)\text{ABCD is a rectangle}\text{Proof :In parallelogram PQRS,}PS||QR\left(\text{opposite sides}\right)\angle P+\angle Q={180}^{\circ }\therefore \text{and AP and AQ are the bisectors of consecutive angle}anglePand\angle Q\text{of the parallelogram}.\therefore \angle APQ+\angle AQP=\frac{1}{2}\times {180}^{\circ }={90}^{\circ }Butin\Delta APQ,\angle A+\angle APQ+\angle AQP={180}^{\circ }\left(\text{Angles of a triangle}\right)\Rightarrow \angle A+{90}^{\circ }={180}^{\circ }\Rightarrow \angle A={180}^{\circ }-{90}^{\circ }\left(v\right)\angle A={90}^{\circ }SimilarlyPQ||SR\left(i\right)\therefore \angle PSB+SPB={90}^{\circ }\left(ii\right)and\angle PBS={90}^{\circ }But,\angle ABC=\angle PBS\left(\text{Vertically opposite angles}\right)\left(iii\right)\therefore \angle ABC={90}^{\circ }\text{Similarly we can prove that}\left(iv\right)\angle ADC={90}^{\circ }and\angle C={90}^{\circ }\left(v\right)\therefore \text{ABCD is a rectangle}\left(\text{Each angle of a quadrilateral is}{90}^{\circ }\right)\text{Hence proved.}$