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 Fig. 7.20). Show that: 

(i) ∆ APB ≅ ∆ AQB 

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

Answer 1

Given: 

In triangles ∆APB and ∆AQB:

• ∠APB = ∠AQB (Each 90º)
• ∠PAB = ∠QAB (l is the angle bisector of ∠A)
• AB = AB (Common side)

To Prove:

By the AAS (Angle-Angle-Side) congruence rule, we can conclude:

\[ \angle \Delta APB \cong \angle \Delta AQB \]

Therefore, by CPCT (Corresponding Parts of Congruent Triangles), we have:

\[ BP = BQ \]

This proves that point B is equidistant from the points A and B. Hence, it can be said that B lies on the perpendicular bisector of line segment AB.