ABC is a triangle right angled at C and M is mid-point of hypotenuse AB. Line drawn through M and parallel to BC intersects AC at D. Show that :

(i) D is mid-point of AC.

(ii) MD is perpendicular to AC.

(iii) CM = MA = $\dfrac{1}{2}$ AB

(i) Given:ABC is a triangle right angled at C and M is mid-point of hypotenuse AB. DM is drawn parallel to the side BC.

To prove: D is mid-point of AC, i.e., AD = DC.

Construction: Draw BF parallel to DC which meets DM produced at F.

Proof: As DF ∥ CB and DC ∥ FB. So, CBFD is a parallelogram.

⇒ BF = DC (opposite sides of a parallelogram are equal.)

In △ ADM and △ BFM,

AM = MB (M is mid-point of AB)

∠AMD = ∠BMF (Vertically opposite angles)

∠MAD = ∠MBF (Alternate angles)

Using ASA congruency criterion,

△ ADM ≅ △ BFM

By corresponding parts of congruent triangles,

⇒ AD = FB

⇒ AD = DC (As FB = DC)

Hence, D is mid-point of AC.

(ii) Since DM ∥ BC and AC is traversal, we can use the consecutive angles between parallel lines.

⇒ ∠MDC + ∠DCB = 180° (Linear pair of angles)

⇒ ∠MDC + 90° = 180°

⇒ ∠MDC = 180° - 90°

⇒ ∠MDC = 90°

Hence, MD is perpendicular to AC.

(iii) As proved above,

∠ADM = ∠MDC = 90°

Also, D is mid-point of AC and DM is parallel to CB; DM bisects side AC, i.e., AD = DC.

In △ AMD and △ CMD.

AD = CD (Proved above)

∠ADM = ∠CDM (Both are 90°)

DM = DM (Common)

By SAS congruency criterion,

△ AMD ≅ △ CMD

By corresponding parts of congruent triangles,

CM = AM = $\dfrac{1}{2}$ AB

Hence, CM = MA = $\dfrac{1}{2}$ AB.