Question

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

Answer 1

R.E.F. Image.

In $△ABC,$

(1) M is the mid point of AB and

$MD\parallel BC$

D is the mid point of AC

(2) As $MD\parallel BC$ &

AC is transversal

$\angle MDC+\angle BCD={180}^{\circ }$

$\angle MDC+{90}^{\circ }={180}^{\circ }\Rightarrow \angle MDC={90}^{\circ }\Rightarrow MD\perp AC$

(3) In $△AMD$ and $△CMD$

$AD=CD$

$\angle ADM=\angle CDM$

$DM=DM$

$△AMD\cong △CMD$

$AM=CM\Rightarrow AM=\frac{1}{2}AB\Rightarrow CM=AM=\frac{1}{2}AB$