Question

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of angle A. Show that : 1. Triangle APB congruent to triangle AQB 2. BP equals to BQ or B is equidistant from the arms of angle A.
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Answer 1

SOLUTION:-

Given:-

• Line l is the bisector of an angle A and B is any point on l.
• BP and BQ are perpendiculars from B to the arms of angle A.

Need to show:-

• ∆APB congruent to ∆AQB.
• BP = BQ or B is equidistant from the arms of <A.

Step by step explanation:-

In ∆ APB and ∆ AQB,

=> <P = <Q [Right angles]

=> <BAP = <BAQ [l is the bisector]

=> AB = AB [common]

From AAS congruency,

=> ∆ APB congruent to ∆ AQB

Hence proved✔️

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As we had proved above that,

=> ∆ APB congruent to ∆ AQB

Therefore,

=> BP = BQ [By CPCT]

Therefore,

B is equidistant from the arms of <A.

Hence proved✔️

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Answer 2

Answer:

Given: line L is bisector of angle A and B is any point on L. BP and BQ are perpendicular from B to arm of angle A

Step-by-step explanation:

1) In triangle APB and AQB

angle BAP=BAQ

AB=AB(COMMON)

ANGLE BPA=BQA

2) TRIANGLE APB IS CONGRANT to TRIANGLE AQB( AAS)

BP=BQ (CPCT)