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எங்கள் ஆசிரியர்களுடன் 1-ஆன்-1 ஆலோசனை நேரத்தைப் பெறுங்கள். டாப்பர் ஆவதற்கு நாங்கள் பயிற்சி அளிப்போம்

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