UPSKILL MATH PLUS

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