Question

ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallels to BC intersect AC at D. Show that CM=MA=$\frac{1}{2}AB$

Answer 1

To show : $CM=MA=\frac{1}{2}AB$

In $△AMD$ and $△CMD$

$AD=DC$ ($D$ is midpoint)

$\angle ADM=\angle CDM$

$DM=DM$ (common)

$△AMD=\cong △CMD$ ($SAS$ congruence rule )

$\therefore AM=CM\left(CPCT\right)-------------\left(1\right)$

$AM=\frac{1}{2}AB\left(Mismidpoint\right)----------\left(2\right)$

from eq $\left(1\right)$ and $\left(2\right)$

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