In right triangle ABC, right angled …

Given: 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.
To Prove:

1. $\triangle$AMC $\cong$ $\triangle$BMD
2. $\angle$DBC is a right angle
3. $\triangle$DBC $\cong$ $\triangle$ACB
4. CM = $\frac {1} {2}$AB

Proof:

1. In $\triangle$AMC and $\triangle$BMD
   AM = BM ...[As M is the mid-point]
   CM = DM ...[Given]
   $\angle$AMC = $\angle$BMD ...[Vertically opposite angles]
   $\therefore$ $\triangle$AMC $\cong$ $\triangle$BMD proved ...[SAS property] ...(1)
2. $\triangle$AMC $\cong$ $\triangle$BMD ...[From (1)]
   $\angle$ACM = $\angle$BDM ...[c.p.c.t.]
   These are alternate interior angles and they are equal.
   $\therefore$ AC $\|$ BD
   As AC $\|$ BD and transversal BC intersects them
   $\therefore$ $\angle$DBC + $\angle$ACB = 180° ...[Sum of the consecutive interior angles of the transversal]
   $\angle$DBC + 90° = 180°
   $\angle$DBC = 180° - 90° = 90°
   $\angle$DBC is a right angle proved.
3. $\triangle$AMC $\cong$ $\triangle$BMD ...[From (1)]
   $\therefore$ AC = BD ...[c.p.c.t.] ...(2)
   In DDBC and DACB
   BC = CB ...[Common]
   $\angle$DBC = $\angle$ACB ...[each = 90° as proved above]
   BD = CA ...[From (2)]
   $\therefore$ $\triangle$DBC $\cong$ $\triangle$ACB ...[SAS property]
4. DDBC $\cong$ DACB ...[As proved in (iii)]
   $\therefore$ DC = AB ...[c.p.c.t.]
   $\therefore$ 2CM = AB ...[DM = CM = $\frac {1} {2}$ DC]
   $\therefore$ CM =$\frac {1} {2}$ AB