Answer variants:
altitude
CPCT
two
\(\angle AXB = \angle AYC\)
common
\(\triangle ACY\) and \(\triangle ABX\)
\(CY\) is perpendicular to \(AB\)
\(BX\) and \(CY\)
\(AC\)
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
is a perpendicular line segment drawn through the vertex 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, corresponding pairs of angles and one 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.