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8.(iv) In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that \small DM=CM. Point D is joined to point B (see Fig.). Show that:

  (iv)   \small CM=\frac{1}{2}AB


Answers (1)


In the previous part we have proved that  \Delta DBC\ \cong \ \Delta ACB.

Thus by c.p.c.t., we can write :        DC\ =\ AB

                                     DM\ +\ CM\ =\ AM\ +\ BM

or                                  CM\ +\ CM\ =\ AB                                              (Since M is midpoint.)

or                                                      \small CM=\frac{1}{2}AB.

Hence proved.             

Posted by

Devendra Khairwa

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