In a triangle \(ABC\), \(D\) is the mid-point of side \(AC\) such that \(BD = \frac{1}{2} AC\). Show that \(\angle ABC\) is a right angle.
 
Proof:
 
Given that in \(\triangle ABC\), \(D\) is the mid-point of side \(AC\) such that \(BD = \frac{1}{2} AC\).
 
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\(AD = \) [Since \(D\) is the midpoint]
 
Now, \(AC = AD + DC\)
 
\(AC = AD + AD\) or \(AC = DC + DC\)
 
\(AC = 2AD\) or \(AC = 2DC\)
 
\(AD = DC = \frac{1}{2} AC\) ---- (\(1\))
 
\(BD = \frac{1}{2} AC\) ---- (\(2\)) [Given]
 
From (\(1\)) and (\(2\)), we have:
 
\(AD =\) \(= BD\) ---- (\(3\))
 
In \(\triangle DAB\), 
 
\(AD = BD\) [From (\(3\))]
 
We know that, the angles opposite to sides are equal
 
\(\angle DAB = \angle DBA\) ---- (\(4\))
 
Similarly, in \(\triangle DCB\),
 
\(BD = DC\) [From (\(3\))]
 
\(\angle DBC = \angle \) ---- (\(5\)) [Angles opposite to equal sides are equal]
 
Applying angle sum property in \(\triangle ABC\), we have:
 
\(\angle ABC + \angle BAC + \angle ACB = 180^{\circ}\)
 
\(\angle ABC + \angle BAD + \angle DCB = 180^{\circ}\) [From the figure \(\angle BAC = \angle BAD\) and \(\angle ACB = \angle DCB\)]
 
\(\angle ABC + \angle DBA + \angle DBC = 180^{\circ}\) [Using (\(4\)) and (\(5\))]
 
\(\angle ABC + \angle ABC = 180^{\circ}\) [Since \(\angle ABD + \angle DBC = \angle ABC\)]
 
\(2 \angle ABC = 180^{\circ}\)
 
\(\angle ABC = \)\(^{\circ}\)
 
Therefore, \(\angle ABC\) is a right angle.
 
Hence, we proved.