Question

plz solve this question

Answer 1

soln
for ∆APB and ∆AQB,
hypo(AB) congruent hypo(AB) ......( common side )
angle AQB congruent angle APB .....( each ninety )
angle QAB congruent angle BAP ......( angle l is the bisector of angle A)
therefore the triangles are congruent by hypotenuse side theorem
I.e ∆APB congruent ∆AQB
therefore PB = QB ....(c.s.c.t)
i.e BP=BQ
hence proved .
.

Answer 2

 ★★★ Hiii friend...!!  ★★★



Here is the answer....!!



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i) In Δ APB and  Δ AQB we have :



   →  ∠ BAP = ∠ BAQ ( it is given that AB is a bisector.....)                      


   →  AB = AB                             ( Common..)                                       

   → ∠ APB = ∠ AQB                 ( Both 90 degree angles....)                 


∴ →  Δ APB  = Δ AQB               (  by AAS congruence rule....)              


Hence proved.....!!



ii) From solution (i), we know that 


→   Δ APB  = Δ AQB        


∴ BP=BQ       ( By C.P.C.T..... Corresponding parts of congruent triangles are equal)


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 ★★★ I hope this may help u....!! ★★★