In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Proof:
In \(\triangle ADB\) and \(\triangle CDE\), we have:
\(AD = DC\) [Since \(D\) is the midpoint]
\(BD = DE\) [Since \(D\) is the midpoint]
\(\angle ADB = \angle \) [Vertically opposite angles]
Then, by congruence rule\(\triangle ADB \cong \triangle CDE\)
So, \(AB = EC\) ---- (\(1\)) [By CPCT]
\(\angle BAD = \angle DCE\) [Alternate angles]
Now, \(\angle ABC + \angle BCE =\) \(^{\circ}\) [Cointerior angles]
\(90^{\circ} + \angle BCE = 180^{\circ}\)
\(\angle BCE = 180^{\circ} - 90^{\circ}\)
\(\angle BCE = 90^{\circ}\)
We know that \(BD = \).
\(BE = BD + DE\)
\(BE = BD + BD\)
\(BE = 2BD\)
\(BD = \frac{1}{2}BE\) ---- (\(2\))
In \(\triangle ABC\) and \(\triangle ECB\), we have:
\(AB = EC\) [using (\(1\))]
\(BC = BC\) [Common side]
\(\angle ABC = \angle ECB = 90^{\circ}\)
Then, by congruence rule, \(\triangle ABC \cong \triangle ECB\)
So, \(AC = EB\) [By CPCT]
\(\frac{1}{2}AC = \frac{1}{2}EB\) [Dividing by \(2\) on both sides]
\(\frac{1}{2}AC = BD\) [Using (\(2\))]
Therefore, the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.
Hence, we proved.