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)    ∆AMC ≅ ∆BMD
(ii)    ∠DBC is a right angle
(iii)    ∆DBC ≅ ∆ACB

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: (i) ∆AMC ≅ ∆BMD
(ii)    ∠DBC is a right angle
(iii)    ∠DBC ≅ ∆ACB

(i) In ∆AMC and ∆BMD,
AM = BM
| ∵ M is the mid-point of the hypotenuse AB
CM = DM    | Given
∠AMC = ∠BMD
| Vertically Opposite Angles
∴ ∆AMC ≅ ∆BMD. | SAS Rule
(ii)    ∵ ∆AMC ≅ ∆BMD
| From (i) above
∠ACM = ∠BDM    | C.P.C.T.
But these are alternate interior angles and they are equal
∴ AC || BD
Now, AC || BD and a transversal BC intersects them
∴ ∠DBC + ∠ACB = 180°
| ∵ The sum of the consecutive interior angles on the same side of a transversal is
180°
⇒ ∠DBC + 90° = 180°
| ∵ ∠ACB = 90° (given)
⇒     ∠DBC = 180° - 90° = 90°
⇒ ∠DBC is a right angle.
(iii)    In ∆DBC and ∆ACB,
∠DBC = ∠ACB (each = 90°)
| Proved in (ii) above
BC = CB    | Common
∵ ∆AMC ≅ ∆BMD    | Proved in (i) above
∴ AC = BD    | C.P.C.T.
∴ ∆DBC ≅ ∆ACB.    | SAS Rule
(iv)    ∵ ∆DBC ≅ ∆ACB
| Proved in (iii) above
∴ DC = AB    | C.P.C.T.
      2CM = AB