Question

Line is the bisector of an angle angle A and B is any point on . BP and BQ are perpendiculars from B to the arms of angle A .Show that
i) triangle APB =/ triangle AQB
ii) B is equidistant from the arms of
Angle A

Answer 1

In △APB and △AQB

∠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.

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