Question

Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms ot ∠A.

Answer 1

Answer:

In △APB and △AQB

In △APB and △AQB∠APB=∠AQB (Each 90

In △APB and △AQB∠APB=∠AQB (Each 90 o

In △APB and △AQB∠APB=∠AQB (Each 90 o )

In △APB and △AQB∠APB=∠AQB (Each 90 o )∠PAB=∠QAB (l is the angle bisector of ∠A)

In △APB and △AQB∠APB=∠AQB (Each 90 o )∠PAB=∠QAB (l is the angle bisector of ∠A)AB=AB (Common)

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)

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)

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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