Question

Bc idorsects AC ad D. show that
;) HDIAC
ii) CH = MA = 1 AB
the mid-boint H of hypotonuse AB and parallel to​

Answer 1

Step-by-step explanation:

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.

Answer 2

Answer:

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