# Ex.7.3 Q2 Triangles Solution - NCERT Maths Class 9

## Question

\(AD\) is an altitude of an isosceles triangle

\(ABC\) in which \(AB = AC.\) Show that

(i) \(AD\) bisects \(BC\)

(ii) \(AD\) bisects \(\angle A\).

## Text Solution

**What is Known?**

AD is an altitude of an isosceles triangle \(ABC\) in which \(AB\,\, AC.\)

**To prove:**

i) \(AD\) bisects \(BC\)

ii) \(AD\) bisects

**Reasoning:**

We can show triangle \(BAD\) and \(CAD\) congruent by using RHS congruency criterion and then we can say corresponding parts of congruent triangle are equal.

**Steps:**

(i) In \(\Delta BAD\) and \(\Delta CAD\),

\[\begin{align} \angle ADB &= \angle ADC\\(\text{Each }90^{\circ}&\text{ as }AD\text{ is the altitude})\\\\AB &= AC\;(\text{Given})\\AD &= AD\;(\text{Common})\\ \therefore \Delta BAD &\cong \Delta CAD\\(\text{By RHS }&\text{ Congruence rule})\end{align}\]

\(\therefore BD = CD\) (By \(CPCT)\)

Hence, \(AD\) bisects \(BC\).

(ii) Also, by \(CPCT\),

\(\angle BAD = \angle CAD\)

Hence, \(AD\) bisects \(\angle A\).