# Ex.7.1 Q5 Triangles Solution - NCERT Maths Class 9

## Question

Line \(l\) is the bisector of an angle \(\angle A\) and \(\angle B\) is any point on \(l\). \(BP\) and \(BQ\) are perpendiculars from \(B\) to the arms of \(\angle A\) (see the given figure).

Show that:

(i) \(\Delta APB \cong \Delta AQB\)

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

## Text Solution

**What is known?**

\(l\) is the bisector of an angle \(\angle {\text{A}}\) and \(BP \bot AP\;{\text{and BQ}} \bot {\text{AQ}}\)

**To prove:**

\(\Delta {\text{APB}} \cong \Delta {\text{AQB}}\) and \({\text{BP}}\,{\text{ = }}\,{\text{BQ}}\) or B is equidistant from the arms of \(\angle {\text{A}}\)

**Reasoning:**

We can show two triangles APB and AQB congruent by using AAS congruency rule and then we can say corresponding parts of congruent triangles will be equal.

**Steps:**

In \(\Delta APB\) and \(\Delta AQB\),

\(\begin{align}&\angle BAP= \angle BAQ \\&\text{(l) is the angle bisector of }(\angle A)\\\\&\angle APB = AQB ( \text{Each} \;90^ {\circ})\\AB&=AB \text{(Common)}\\&\therefore \Delta APB \cong \Delta AQB \\&\text{(By AAS congruence rule)}\\\\\therefore BP &= BQ (By CPCT)\end{align}\)

Or, it can be said that \(B \) is equidistant from the arms of \(\angle A\).