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Question 5 ii...

Question

Question 5 (ii)
Line l is the bisector of an angle $∠ A a n d B$ is any point on l. BP and BQ are perpendiculars from B to the arms of $∠ A$ (see the given figure). Show that:
BP = BQ or B is equidistant from the arms of $∠ A .$

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Solution

$Δ A P B a n d Δ A Q B ∠ A P B = ∠ A Q B (E a c h 90^{\circ}) ∠ P A B = ∠ Q A B (I i s a n g l e b i s e c t o r o f ∠ A) A B = A B (C o m m o n) ∴ Δ A P B ≅ Δ A Q B (B y A A S C o n g r u e n c e R u l e) ∴ B P = B Q (B y C P C T)$

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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 the given figure). Show that:

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