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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 (iii) CM = MA = 1/2 AB

Q: 7 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 \small CM=MA=\frac{1}{2}AB
 

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Given: 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.

To prove :  \small CM=MA=\frac{1}{2}AB

Proof : In \triangleABC,     

 M is the midpoint of AB.   (Given)

 DM || BC            (Given)

 By converse of midpoint theorem,

 D is the midpoint of AC  i.e.  AD = DC.

In \triangle AMD  and \triangle CMD,

  AD = DC    (proved above)

   \angleADM = \angle CDM  (Each  right angle)

  DM = DM      (Common)

  \triangle AMD \cong\triangle CMD (By SAS)

  AM = CM         (CPCT)

  But ,\small AM=\frac{1}{2}AB

  Hence,\small CM=MA=\frac{1}{2}AB

 

 

 

     

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