Fewpal
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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) D is the mid-point of AC. 

(ii) MD ⊥ AC 

(iii) CM = MA = 1/ 2 AB

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Given : A triangle ABC, in which ∠C = 90° and M is the mid-point of AB and BC || DM. 

To Prove : 

(i) D is the mid-point of AC [Given] 

(ii) DM ⊥ BC 

(iii) CM = MA = 1/ 2 AB 

Construction : Join CM. 

Proof : 

(i) In ∆ABC, 

M is the mid-point of AB. [Given] 

BC || DM [Given] 

D is the mid-point of AC [Converse of mid-point theorem] Proved. 

(ii) ∠ADM = ∠ACB [∵ Coresponding angles] 

But ∠ACB = 90° [Given] 

∴ ∠ADM = 90° 

But ∠ADM + ∠CDM = 180° [Linear pair] 

∴ ∠CDM = 90° 

Hence, MD ⊥ AC Proved. 

(iii) AD = DC …(1) [∵ D is the mid-point of AC] 

Now, in ∆ADM and ∆CMD, we have 

∠ADM = ∠CDM [Each = 90°] 

AD = DC [From (1)] DM = DM [Common] 

∴ ∆ADM ≅ ∆CMD [SAS congruence] 

⇒ CM = MA …(2) [CPCT] 

Since M is mid-point of AB, 

∴ MA = 1/ 2 AB …(3) 

Hence, CM = MA = 1 /2 AB Proved. [From (2) and (3)]

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