\(ABC\) is an isosceles triangle in which \(AC = BC\). \(AD\) and \(BE\) are respectively two altitudes to sides \(BC\) and \(AC\). Prove that \(AE = BD\).
Proof:
In \(\triangle ADC\) and \(\triangle BEC\),
\(AC =\) ---- (\(1\)) [Given]
\(\angle ADC = \angle\) \(= 90^{\circ}\) [Since \(AD\) and \(BE\) are altitudes]
\(\angle ACD = \angle BCE\) [Common angle]
Then, by congruence rule\(\triangle ADC \cong \triangle BEC\)
So, by \(CPCT\), \(CE =\) ---- (\(2\))
Subtracting equation (\(2\)) from (\(1\)), we get:
\(AC - CE = BC - CD\)
\(AE = BD\)