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Answer variants:
\(CY\) is perpendicular to \(AB\)
CPCT
altitudes
perpendicular line segment
two
\(\triangle ACY\)
\(\angle AXB = \angle AYC\)
\(BX\)
\(ACY\)
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
An altitude is a perpendicular line segment drawn through the vertex of the triangle to the opposite side.
Here, \(CY\) is an altitude of \(AB\), and is an altitude of \(AC\).
Hence, and \(BX\) is perpendicular to \(AC\).
To prove that the altitudes are equal, let us consider and \(\triangle ABX\).
Here, \(AB = AC\) [Given]
Also, as the altitudes meet the sides at right angles.
Also, \(\angle A\) is common to both triangles 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.
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