pa is perpendicular to ab , qb is perpendicular to ab .and pa=qb.prove triangle oap congruent to triangle obq.is oa =op
Answer :
From given information we form our diagram , as , and we take point " O " where line PB and QA intersect .
Now In PAB and QBA
PA = QB ( Given )
PAB = QBA = 90 ( Given )
And
AB = AB ( Common side )
Hence
PAB QBA ( By SAS rule )
So,
APB = BQA ( By CPCT ) --------------- ( 1 )
And
PB = QA ( By CPCT ) --------------- ( 2 )
Now In OAP and OBQ
PA = QB ( Given )
APO = BQO ( From equation 1 ) ( Here APB and APO are same angles and BQA and BQO are same angles )
And
AOP = BOQ ( Vertically opposite angles )
Hence
OAP OBQ ( By AAS rule ) ( Hence proved )
So,
OA = OB ( By CPCT ) --------------- ( 3 )
OP = OQ ( By CPCT ) --------------- ( 4 )
So. from equation 2 ,3 and 4 , we get
OA = OP ( Hence proved )
From given information we form our diagram , as , and we take point " O " where line PB and QA intersect .
Now In PAB and QBA
PA = QB ( Given )
PAB = QBA = 90 ( Given )
And
AB = AB ( Common side )
Hence
PAB QBA ( By SAS rule )
So,
APB = BQA ( By CPCT ) --------------- ( 1 )
And
PB = QA ( By CPCT ) --------------- ( 2 )
Now In OAP and OBQ
PA = QB ( Given )
APO = BQO ( From equation 1 ) ( Here APB and APO are same angles and BQA and BQO are same angles )
And
AOP = BOQ ( Vertically opposite angles )
Hence
OAP OBQ ( By AAS rule ) ( Hence proved )
So,
OA = OB ( By CPCT ) --------------- ( 3 )
OP = OQ ( By CPCT ) --------------- ( 4 )
So. from equation 2 ,3 and 4 , we get
OA = OP ( Hence proved )