Please solve this problem
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for me it did not got answer option
so i gave this answer as comment
thank you so much @nikithgandhivalaval
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Step-by-step explanation:
given:
l is the bisector of < A
so,<PAB= <QAB ....(1)
BP and BQ are perpendicular from B,
So,<APB = <AQB = 90° ......(2)
To prove (i) /\ APB ~= /\ AQB (ii) BP=BQ
Proof:
In /\ APB and /\ AQB,
<APB = < AQB (From (2))
< PAB = < QAB (From (1))
AB=AB ( common)
/\ APB ~= /\ AQB ( AAS congruence rule )
BP = BQ (CPCT)
Hence proved
Similar questions
∠APB=∠AQB (Each 90
o
)
∠PAB=∠QAB (l is the angle bisector of ∠A)
AB=AB (Common)
∴△APB≅△AQB (By AAS congruence rule)
∴BP=BQ (By CPCT)
It can be said that B is equidistant from the arms of ∠A.