Answer variants:
opposite side
corresponding pair of sides
right angles
\(BX\) is perpendicular to \(AC\)
\(ACY\)
altitudes
\(\triangle ACY\) and \(\triangle ABX\)
\(BX\)
CPCT
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 \(AB\), and 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, \(\angle AXB = \angle AYC\), 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 \(\triangle ACY \cong \triangle ABX\), and by the altitudes \(CY\) and \(BX\) are equal.