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