**ABC is a triangle right-angled at C. A line through the mid-point M of hypotenuse AB parallel to BC intersects AC … AC (iii) CM =…**

###### Answers:

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.