# Ex.7.4 Q2 Triangles Solution - NCERT Maths Class 9

## Question

In the given figure sides \(AB\) and \(AC\) of \(\Delta ABC\) are extended to points \(P\) and \(Q\) respectively. Also, \(\angle PBC \lt \angle QCB\). Show that \(AC \gt \angle AB\).

## Text Solution

**What is Known?**

\(\angle \text{PBC}<\angle \text{QCB}\text{.}\)

**To prove:**

\(\text{AC}>\text{AB}\text{.}\)

**Reasoning:**

By using linear pair, we can find inequality of interior angles and then we can use the fact that In any triangle, the side opposite to the larger (greater) angle is longer.

**Steps:**

In the given figure,

\[\begin{align} &\angle ABC +\angle PBC=180^{\circ}\,\,\,\\&\text{( Linear pair)} \\\\& \angle ABC =180^{\circ} - \angle PBC\, \ldots (1) \\ \end{align}\]

Also,

\[\begin{align} &\angle ACB +\angle QCB=180^{\circ} \\ &\angle ACB =180^{\circ} - \angle QCB\, \ldots (2) \\ \end{align}\]

As \(\angle PBC \lt \angle QCB\),

\[\begin{align} & 180^{\circ} - PBC \gt 180^{\circ} - \angle QCB\\ & \angle ABC \gt ACB\\&[\text{From Equations (1) and (2)}]\\\\ & AC \gt AB \\&\left (\begin{array}\ \text{side opposite to the}\\\text{ larger angle is larger.}\end{array} \right)\end{align}\]

Hence proved, \(AC \gt AB\)