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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 ∠ A (see Figure). Show that:
(i) Triangle APB ≅ triangle AQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A.

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(i) ∠APB = ∠AQB (It is given that both the angles are 90°) (ii) In ΔABP and ΔABQ, ∠APB = ∠AQB (both are 90°) ∠BAP = ∠BAQ (line l bisects ∠A into ∠BAP and ∠BAQ) Side...
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(i) ∠APB = ∠AQB (It is given that both the angles are 90°)

(ii) In ΔABP and ΔABQ, ∠APB = ∠AQB (both are 90°)

                                         ∠BAP = ∠BAQ (line l bisects ∠A into ∠BAP and ∠BAQ)

                                         Side BA ≅ Side BA (common side of both the triangles)

Hence, ΔABP ≅ ΔABQ by AAS Test.

Hence, BP = BQ (corressponding sides of congruent triangles)

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∠BAP =∠BAQ : given ∠ BPA = ∠BQA : both are 90 degree. Given. So third angles of both triangles be also equal. So ∠ ABQ = ∠ ABP Now as per ASA congruency requirement: As ∠QBA, side AB, ∠ QAB = ∠PBA, side AB, ∠BAP So both Δ ABQ ≅ Δ ABP. And when...
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∠BAP =∠BAQ : given

∠ BPA = ∠BQA : both are 90 degree. Given.

So third angles of both triangles be also equal.

So ∠ ABQ = ∠ ABP

Now as per ASA congruency requirement:

As ∠QBA, side AB, ∠ QAB = ∠PBA, side AB, ∠BAP

So both Δ ABQ ≅ Δ ABP.

 

And when all angles are equal, and a common side AB is equal for both triangles, so other two sides will be equal too. 

So side BP = side BQ

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