Answer variants:
\(\triangle ACY \cong \triangle ABX\)
\(CY\) is perpendicular to \(AB\)
\(\triangle ABX\)
\(BX\) and \(CY\)
\(\angle A\)
opposite side
\(\angle AXB = \angle AYC\)
\(CY\)
corresponding pair of sides
Prove that the altitudes are \(BX\) and \(CY\) are equal if triangle \(ABC\) is isosceles with \(AB = AC\).
Proof:
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, is an altitude of \(AB\), and \(BX\) is an altitude of \(AC\).
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, 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 one are equal.
Thus by congruence criterion, \(\triangle ACY \cong \triangle ABX\).
Since and by CPCT, the altitudes \(CY\) and \(BX\) are equal.