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.