Question

ABC is a triangle right angled at C. A line through the mid point M of the hypotenuse AB and parallel to BC, intersects AC at D. Show that,
(i) D is the mid-point of AC.
(ii) $MD\perp AC$
(iii) $CM=MA=\frac{1}{2}AB$

Answer 1

(i) In $△BAC,$
M is the mid-point of AB and MD || BC.
D is the mid-point of AC.
(Converse of mid-point theorem)
$\therefore AD=CD$

(ii) $\angle ACB=\angle ADM$ (Corresponding angles)
Also, $\angle ACB=90°$
$\Rightarrow ADM=90°$ and $MD\perp AC$

(iii) Consider $△AMD$ and $△CMD$
$DM=DM$(Common side)
$\angle ADM=\angle CDM$ (Right angle)
$AD=CD$ (D is the mid-point of side AC)
So, by SAS congruence criterion, $△AMD\cong △CMD$
$\therefore CM=MA=\frac{1}{2}AB$