\(AD\) is an altitude of an isosceles triangle \(ABC\) in which \(AB = AC\).
Show that:
(i) \(AD\) bisects \(BC\).
(ii) \(AD\) bisects \(∠A\).
Proof:
(i) In \(∆ADB\) and \(∆ADC\)
\(∠ADB=∠\) [Right angles]
\(AB = AC\) [Given]
\(AD = AD\) [Common side]
Therefore, by rule \(∆ADB ≅∆ADC\)
\(BD = \) [By CPCT]
Therefore, \(AD\) bisects \(BC\).
(ii) Since \(∆ADB ≅∆ADC\) [By SAS rule]
\(\angle BAD = \angle \) [By CPCT]
Therefore, \(AD\) bisects \(∠A\).