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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

              (iii)   \small CM=MA=\frac{1}{2}AB
 

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M mansi

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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