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Answer variants:
right angles
corresponding pair of sides
opposite side
\(CY\) is perpendicular to \(AB\)
\(\triangle ABX\)
\(BX\) and \(CY\)
\(\triangle ACY \cong \triangle ABX\)
Prove that the altitudes are \(BX\) and \(CY\) are equal if triangle \(ABC\) is isosceles with \(AB = AC\).
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, \(CY\) is an altitude of \(AB\), and \(BX\) is an altitude of 
and \(BX\) is perpendicular to \(AC\).
To prove that the altitudes are equal, let us consider \(\triangle ACY\) and 
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, \(\triangle ACY \cong \triangle ABX\).
and by CPCT, the altitudes \(CY\) and \(BX\) are equal.
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