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hi..
In fig.8.21 ,
1. In ∆APB and ∆CQD ,
angle APB =angle CQD (given,it's perpendicular)
DC=AB(opposite sides of parallelogram)
PB= QD (common,on the same side of diagonal)
therefore, ∆APB is congruent to ∆CQD (SAS)
2.AP=CQ(CPCT)
Hope it helps you
In fig.8.21 ,
1. In ∆APB and ∆CQD ,
angle APB =angle CQD (given,it's perpendicular)
DC=AB(opposite sides of parallelogram)
PB= QD (common,on the same side of diagonal)
therefore, ∆APB is congruent to ∆CQD (SAS)
2.AP=CQ(CPCT)
Hope it helps you
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Answer: hope this help you please mark as a brainlist
Step-by-step explanation:(i) In ∆APB and ∆CQD, we have∠APB = ∠CQD [Each 90°]AB = CD [ ∵ Opposite sides of a parallelogram ABCD are equal]∠ABP = ∠CDQ[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]∴ ∆APB = ∆CQD [By AAS congruency](ii) Since, ∆APB ≅ ∆CQD [Proved]⇒ AP = CQ [By C.P.C.T.]
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