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

(i) In ΔABC,
M is mid-point to AB [∵ Given]
and DM ∥ BC [∵ Given]
Hence, D is mid-point of AC [∵ Converse of Mid-point Theorem]

(ii) ∠ADM = ∠ACB [∵ Corresponding angles]
⇒ ∠ADM = 90° [∵ ∠ACB 90°]
Hence, MD ⊥ AC

(iii) In ΔAMD = ΔCMD,
AD = DC [∵ Proved above]
∠ADM = ∠CDM [∵ Each 90°]
DM = DM [∵ Common]
Hence, ΔAMD ΔCMD [∵ SAS Congruency rule]
AM = CM [∵ CPCT]
But AM = (1/2) AB [∵ Given]
Therefore, CM = AM = (1/2) AB