Question

Linel is the bisector of an angle 2 A and B is any
point on I. BP and BQ are perpendiculars from B
to the arms of ZA (see Fig. 7.20). Show that:
(1) A APB=AAQB
(2) BP = BQ or B is equidistant from the arms
of angle A​

Answer 1

$\small \purple {\sf{quєѕtíσn !}}$

Linel is the bisector of an angle 2 A and B is any

point on I. BP and BQ are perpendiculars from B

to the arms of ZA (see Fig. 7.20). Show that:

(1) A APB=AAQB

(2) BP = BQ or B is equidistant from the arms

of angle A

$\small \blue {\sf{αnѕwєr :- }}$

In △s APB and ABQ, we have

∠APB=∠AQB (Each 90 ∘ )

∠PAB=∠QAB (AB bisect ∠PAQ)

AB=BA (common)

Therefore, △APB≅△ABQ (AAS)

⇒ BP=BQ (cpct)

Hence, B is equidistant from the anus of ∠A.

Answer 2

Answer:

Linel is the bisector of an angle 2 A and B is any

point on I. BP and BQ are perpendiculars from B

to the arms of ZA (see Fig. 7.20). Show that:

(1) A APB=AAQB

(2) BP = BQ or B is equidistant from the arms

of angle A

Answer:-In △s APB and ABQ, we have

∠APB=∠AQB (Each 90 ∘ )

∠PAB=∠QAB (AB bisect ∠PAQ)

AB=BA (common)

Therefore, △APB≅△ABQ (AAS)

⇒ BP=BQ (cpct)

Hence, B is equidistant from the anus of ∠A.