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

Given: \(\triangle{ABC}\) is a right angled triangle \(\Rightarrow \) \(\angle{C}\) = \(90^\circ\)
and M is the mid-point of AB.
Also, DM || BC

i) In \(\triangle{ABC}\), M is the mid-point of AB and BC || MD,

So, By converse of mid-point theorem,
D is the mid-point of AC.

\(\Rightarrow \) AD = CD ...(i)


ii) Since, BC || MD and CD is transversal.
\(\therefore \) \(\angle{ADM}\) = \(\angle{ACB}\)
(Corresponding angles)
But, \(\angle{C}\) = \(90^\circ\)
\(\therefore \) \(\angle{ADM}\) = \(90^\circ\)

Hence, proved that MD is perpendicular to AC.


iii) Now, in \(\triangle{ADM}\) and \(\triangle{ACM}\), we have,
AD = CD ...(from (i))
DM = MD ...(Common side)
\(\angle{ADM}\) = \(\angle{MDC}\) = \(90^\circ\) ...(Proved)
\(\therefore \) \(\triangle{ADM}\) \(\displaystyle \cong \) \(\triangle{CDM}\) ...(By SAS rule)
\(\therefore\) CM = AM ...(ii)(By CPCT)

Also, M is the mid point of AB.
\(\Rightarrow \) AM = BM = (\(\frac{1}{2} \) ) AB ...(iii)

Thus, from (ii) and (iii),
CM = AM = (\(\frac{1}{2} \) ) AB
Hence, proved.