how to solve this question plzzzzz hurry
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Since l is the angle bisector,
<QAB = <PAB
So, in Δ AQB and ΔAPB,
<QAB= <PAB (proved above)
AB = AB (common)
<AQB= <APB (each 90)
∴Δ AQB is congruent to ΔAPB.
∴BP=BQ (CPCT), hence B is equidistant from P and Q.
Hope it helped!
NISHTHA.
<QAB = <PAB
So, in Δ AQB and ΔAPB,
<QAB= <PAB (proved above)
AB = AB (common)
<AQB= <APB (each 90)
∴Δ AQB is congruent to ΔAPB.
∴BP=BQ (CPCT), hence B is equidistant from P and Q.
Hope it helped!
NISHTHA.
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pls mark it as a brainlist
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