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 i)angle apb congruent to angle aqb and bp=bq About the author Arya
Step-by-step explanation: 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. Reply
Answer: 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 Reply
Step-by-step explanation:
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.
Answer:
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