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Question

ABC is a triangle right angled at C. A line through the mid point M of the hypotenuse AB and parallel to BC, intersects AC at D. Show that,
(i) D is the mid-point of AC.
(ii) MDAC
(iii) CM=MA=12AB


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Solution

(i) In BAC,
M is the mid-point of AB and MD || BC.
D is the mid-point of AC.
(Converse of mid-point theorem)
AD=CD

(ii) ACB=ADM (Corresponding angles)
Also, ACB=90°
ADM=90° and MDAC

(iii) Consider AMD and CMD
DM=DM(Common side)
ADM=CDM (Right angle)
AD=CD (D is the mid-point of side AC)
So, by SAS congruence criterion, AMDCMD
CM=MA=12AB

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