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Answer :
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.