Question

Question 14
In a right triangle, prove that the line-segment joining the mid-point of the hypotenuse to the opposite vertex is half the hypotenuse.

Answer 1

Given In $\Delta ABC,\angle B={90}^{\circ }$ and D is the mid-point of AC.
Construction: Produce BD to E such that BD = DE and join EC.
To prove: $BD=\frac{1}{2}AC$

Proof:
In $\Delta ADBand\Delta CDE,AD=DC[\because Dismid-pointofAC]$
$BD=DE\left[byconstruction\right]$
$and\angle ADB=\angle CDE$ [vertically opposite angles]
$\therefore \Delta ADB\cong \Delta CDE$ [by SAS congruence rule]
$\Rightarrow AB=EC$ [by CPCT]
and $\angle BAD=\angle DCE$ [by CPCT]
But $\angle BAD$ and $\angle DCE$ are alternate angles.
So, EC || AB and BC is a transversal.
$\therefore \angle ABC+\angle BCE={180}^{\circ }$ [cointerior angles]
$\Rightarrow {90}^{\circ }+\angle BCE={180}^{\circ }[\because \angle ABC={90}^{\circ },given]\Rightarrow \angle BCE={180}^{\circ }-{90}^{\circ }\Rightarrow \angle BCE={90}^{\circ }$
In $\Delta ABCand\Delta ECB,AB=EC$ [proved above]
BC = CB [common side]
and $\angle ABC=\angle ECB$ $[each{90}^{\circ }$ ]
$\therefore \Delta ABC\cong \Delta ECB$ [by SAS congruence rule]
$\Rightarrow AC=EB$ [by CPCT]
$\Rightarrow \frac{1}{2}EB=\frac{1}{2}AC$ [dividing both sides by 2]
$\Rightarrow BD=\frac{1}{2}AC$