Question

Diagonal SQ of a parallelogram PQRS bisect ∠ S. Prove SQ also bisect ∠Q.​

Answer 1

$\displaystyle \setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,1)(1,1)(6,1)\qbezier(1,1)(1,1)(1.6,4)\qbezier(1.6,4)(1.6,4)(6.6,4)\qbezier(6,1)(6,1)(6.6,4)\qbezier(6.6,4)(6.6,4)(1,1)\qbezier(1.6,4)(1.6,4)(6,1)\put(0.7,0.5){\sf P}\put(6,0.5){\sf Q}\put(1.4,4.3){\sf R}\put(6.6,4.3){\sf S}\end{picture}$

To prove that SQ also bisects ∠Q, we need to prove that ∆ PQS and ∆ QRS are congruent.

• QS = QS ( As it is the diagonal )

• ∠ PQS = ∠ SQR

• PS = QR ( Opposite sides of a parallelogram are equal )

Hence , ∆ PQS is congruent to ∆ QRS by S. A. S congruency .

Hence, by CPCT, the diagonal QS also bisects angle Q .

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