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 (SEE FIGURE) Show that Triangle APB =~ Triangle AQB​

Answer 1

Given :

• L is the bisector of ∠A
• BP and BQ are perpendiculars from B to arms of ∠A

To Prove :

• ∆APB ≅ ∆AQB

Solution :

• In ∆APB and ∆AQB ;

→ ∠AQB = ∠APB (90° angles)

→ AB = AB (common)

→ ∠QAB = ∠PAB (L bisector of ∠A)

　∴ ∆APB ≅ ∆AQB (by AAS cong. rule)

Answer 2

$\huge\tt{\boxed{\tt{\red{ANSWER⤵}}}}$

Given,

L is bisector of angle A

Angle QAB=Angle PAB

To prove,

∆APB ≅ ∆AQB

Proof,

In ∆APB and ∆AQB,

AB=AB (common side)

Angle APB=angle AQB (90° degree each)

Angle QAB=Angle PAB (given)

Therefore,

∆APB ≅ ∆AQB by AAS congruence

Hope it helps..

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