Prove that the altitudes are \(BX\) and \(CY\) are equal if triangle \(ABC\) is isosceles with \(AB = AC\).
It is given that are altitudes of triangle \(ABC\).
An altitude is a perpendicular line segment drawn through the vertex of the triangle to the
Here, \(CY\) is an altitude of \(AB\), and \(BX\) is an altitude of .
Hence, and \(BX\) is perpendicular to \(AC\).
To prove that the altitudes are equal, let us consider \(\triangle ACY\) and .
Here, \(AB = AC\) [Given]
Also, \(\angle AXB = \angle AYC\), as the altitudes meet the sides at .
Also, \(\angle A\) is common to both triangles and \(ABX\).
Here, two corresponding pairs of angles and one are equal.
Thus by congruence criterion, \(\triangle ACY \cong \triangle ABX\).
Since and by CPCT, the altitudes \(CY\) and \(BX\) are equal.