In \(∆ ABC\), the bisector \(AX\) of \(∠ A\) is perpendicular to side \(BC\).
Establish that \(AB = AC\) and \(∆ABC\) is isosceles.
Proof:
In \(∆ABX\) and \(∆ACX\),
\(∠ BAX = ∠ \) (Given)
\(AX = AX\) (Common)
\(∠ AXB = ∠ AXC = \)\(^{o}\) (Given)
Then, by \(∆ ABX≅ ∆ ACX\)
So, \(AB = AC\) (By CPCT)
Therefore, \(∆ ABC\) is an isosceles triangle.
Hence, we proved.