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# 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

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.