Given: A △ABC , right - angled at C. A line through the mid - point M of hypotenuse AB parallel to BC intersects AC at D. To Prove: (i) D is the mid - point of AC (ii) MD | AC (iii) CM = MA = 1 / 2 AB. Proof : (i) Since M is the mid point of hyp. AB and MD | | BC . ⇒ D is the mid - point of AC . (ii) Since ∠BCA = 90 ° and MD | | BC [given] ⇒ ∠MDA = ∠BCA = 90 ° [corresp ∠s] ⇒ MD | AC (iii) Now, in △ADM and △CDM MD = MD [common] ∠MDA = ∠MDC [each = 90°] AD = CD [∵ D the mid - point of AC] ⇒ △ADM ≅ △CDM [by SAS congruence axiom] ⇒ AM = CM Also, M is the mid - point of AB [given] ⇒ CM = MA = 1 / 2 = AB.