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# In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see figure). Show that (i) $$\triangle{AMC}$$ $$\displaystyle \cong$$ $$\triangle{BMD}$$ (ii) $$\angle{DBC}$$ is a right angle (iii) $$\triangle{DBC}$$ $$\displaystyle \cong$$ $$\triangle{ACB}$$ (iv)CM = (1/2) AB

Given :
$$\angle{ACB}$$ in which $$\angle{C}$$ = $$90^\circ$$ and M is the mid-point of AB.

To prove :
(i) $$\triangle{AMC}$$ $$\displaystyle \cong$$ $$\triangle{BMD}$$
(ii) $$\angle{DBC}$$ is a right angle
(iii) $$\triangle{DBC}$$ $$\displaystyle \cong$$ $$\triangle{ACB}$$
(iv)CM = ($$\frac{1}{2}$$ ) AB

Construction : Produce CM to D, such that CM = MD. Join DB.

Proof : In $$\triangle{AMC}$$ and $$\triangle{BMD}$$, we have

AM = BM ...(M is the mid-point of AB)
CM = DM ...(Given)
and $$\angle{AMC}$$ = $$\angle{BMD}$$ ...(Vertically opposite angles)
$$\therefore$$ $$\triangle{AMC}$$ $$\displaystyle \cong$$ $$\triangle{BMD}$$(By SAS congruency test)
Hence, part(i) is proved.

Also, AC = DB ...(By CPCT)

$$\angle{1}$$ = $$\angle{2}$$ ...(Alternate angles) and (by CPCT)

$$\therefore$$ BD || CA and BC is transversal.

$$\Rightarrow$$ $$\angle{ACB}$$ + $$\angle{DBC}$$ = $$180^\circ$$
But, $$\angle{ACB}$$ = $$90^\circ$$ ...(given)
$$\Rightarrow$$ $$90^\circ$$ + $$\angle{DBC}$$ = $$180^\circ$$
$$\Rightarrow$$ $$\angle{DBC}$$ = $$180^\circ$$ - $$90^\circ$$
$$\Rightarrow$$ $$\angle{DBC}$$ = $$90^\circ$$

Hence, part(ii) is proved, too.

Now, considering $$\triangle{DBC}$$ and $$\triangle{ACB}$$, we have,

AC = DB ...(from part(i))

Side BC is common.

and $$\angle{DBC}$$ = $$\angle{ACB}$$ = $$90^\circ$$

Therefore, $$\triangle{DBC}$$ $$\displaystyle \cong$$ $$\triangle{ACB}$$ ...(SSA Congruency theorem)
Hence, now, part(iii) is proved, too.

Now, DC = AB ...(By CPCT)

Multipling both sides by $$\frac{1}{2}$$ , we get,
($$\frac{1}{2}$$ ) DC = $$\frac{1}{2}$$ ) AB

Now, as we know, CM = ($$\frac{1}{2}$$ ) DC
Therefore, CM = ($$\frac{1}{2}$$ ) AB
Hence, part (iv) is proved.