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Asked by Sitaram Last Modified
Answer 
Gautam Pal
Maths Teacher for classes 6th to 10th with 6 years experience and having my own YouTube Channel for clearing doubts
(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)
read less∠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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