Answer variants:
\(\angle A\)
one
\(\angle AXB = \angle AYC\)
\(BX\) is perpendicular to \(AC\)
\(\triangle ACY \cong \triangle ABX\)
\(AB\)
opposite side
altitudes
\(\triangle ACY\) and \(\triangle ABX\)
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 vertex of the triangle to the
.
Here, \(CY\) is an altitude of and \(BX\) is an altitude of \(AC\).
Hence, \(CY\) is perpendicular to \(AB\), and .
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, is common 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 and by CPCT, the altitudes \(CY\) and \(BX\) are equal.