Line l is the bisector of an angle ∠A and ∠B is any point on I. BP and BQ are perpendicular from B to the arms of ∠A . Show that
(i)ΔAPB≅ΔAQB
(ii)BP=BQ or B is equidistant from the arms of ∠A.
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Solution
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Given:l is the bisector of ∠A
So, ∠PAB=∠QAB .......(1)
BP and BQ are perpendiculars from B,
So,∠APB=∠AQB=90∘ .......(2)
In △APB and △AQB,
∠APB=∠AQB from (2)
∠PAB=∠QAB from (1)
AB=AB(common)
∴△APB≅△AQB by AAS congruence rule
∴BP=BQ by CPCT
Hence proved.
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(i)ΔAPB≅ΔAQB
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