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

Proof:

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