1. In the following figure, PQRS is a trapezium with PQ||RS. The bisectors of ZP and ZQ meet
at M on side RS such that ZPMQ = 80°. If PM = QM, then show that ZS + ZR = 160°
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⇒ PQRS is trapezium.
⇒ SP∥RQ [ Given ]
We know, sum of two angles between parallel lines is supplementary
∴ ∠QRS+∠RSP=180
o
⇒ ∠QRS+90
o
=180
o
⇒ ∠QRS=180
o
−90
o
=90
o
⇒ ∠PQR+∠QRS+∠RSP+∠SPQ=360
o
⇒ 100
o
+90
o
+90
o
+∠SPQ=360
o
⇒ 280
o
+∠SPQ=360
o
⇒ ∠SPQ=360
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−280
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=80
o
∴ Measures of ∠P and ∠R are 80
o
and 90
o
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