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Question 7 (iii)
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
CM=MA=12AB

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Solution



Consider a right angle ΔABC right angled at C.

C=90

And M is the mid-point of AB. and DM BC

D is the midpoint of AC [By converse of midpoint theorem]

In ΔADM and ΔCDM ,

DM=MD [Common]

AD=CD [Since, D is mid-point of AC]

ADM=CDM [Since \(DM \parallel BC, \angle ADM = 90]

ΔADMΔCDM [SAS congruency]

CM=AM……(i) [CPCT]

AM=BM=12AB..........(ii)

From Equations (i) and (ii) , we get

CM=AM=12AB

Hence , it is proved.




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