Question

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 the below figure). Show that:

(i) ∆ APB ≅ ∆ AQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.​



ANSWER-

Given :
∠PAB = ∠QAB
∠BPA = ∠BQA
To Prove :
Δ APB ≅ Δ AQB
BP = BD
Solution :
In Δ APB and Δ AOB :-

\longmapsto\tt{\angle{PAB}=\angle{QAB}(Given)}⟼∠PAB=∠QAB(Given)

\longmapsto\tt{\angle{BPA}=\angle{BQA}(Each\:90\degree)}⟼∠BPA=∠BQA(Each90°)

\longmapsto\tt{AB=AB(Common)}⟼AB=AB(Common)

So , By ASA Rule ΔAPB ≅ ΔAQB..

Now :
\longmapsto\tt{BP=BQ(By\:CPCT\:Rule)}⟼BP=BQ(ByCPCTRule)

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ASA Rule :
Two Triangles are congruent if two angles and the included side of One triangle is equal to the two angles and the included side of other Triangle.

CPCT Rule :
CPCT means Corresponding Parts of Congruent Triangles.

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Answer 1

Answer:

Yes, the answer you have written after writing the question is absolutely correct