Given : A triangle ABC, in which ∠C = 90° and M is the mid-point of AB and BC || DM.

To Prove :

(i) D is the mid-point of AC [Given]

(ii) DM ⊥ BC

(iii) CM = MA = 1/ 2 AB

**Construction :** Join CM.

**Proof : **

(i) In ∆ABC,

M is the mid-point of AB. [Given]

BC || DM [Given]

D is the mid-point of AC [Converse of mid-point theorem] **Proved. **

(ii) ∠ADM = ∠ACB [∵ Coresponding angles]

But ∠ACB = 90° [Given]

∴ ∠ADM = 90°

But ∠ADM + ∠CDM = 180° [Linear pair]

∴ ∠CDM = 90°

Hence, MD ⊥ AC Proved.

(iii) AD = DC …(1) [∵ D is the mid-point of AC]

Now, in ∆ADM and ∆CMD, we have

∠ADM = ∠CDM [Each = 90°]

AD = DC [From (1)] DM = DM [Common]

∴ ∆ADM ≅ ∆CMD [SAS congruence]

⇒ CM = MA …(2) [CPCT]

Since M is mid-point of AB,

∴ MA = 1/ 2 AB …(3)

Hence, CM = MA = 1 /2 AB **Proved**. [From (2) and (3)]