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Standard VIII

Mathematics

Construction of a Perpendicular Bisector

In the given ...

Question

In the given figure, 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. Show that :
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

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Solution

It is given that the line “l” is the bisector of angle ∠A and the line segments BP and BQ are perpendiculars drawn from l.
(i) ΔAPB and ΔAQB are similar by AAS congruence because:
∠P = ∠Q (They are the two right angles)
AB = AB (It is the common arm)
∠BAP = ∠BAQ (As line l is the bisector of angle A)
So, ΔAPB ≅ ΔAQB.
(ii) By the rule of CPCT, BP = BQ. So, it can be said the point B is equidistant from the arms of ∠A.

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In the given figure, 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. Show that : (i) ΔAPB ≅ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠A.

In the given figure, 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. Show that :
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.