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  • 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 DM = CM. Point D is joined to point B (see Fig). Show that: (i) ∆ AMC ≅ ∆ BMD

8.(i)  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:

 (i) \small \Delta AMC\cong \Delta BMD

                

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Consider \Delta AMC   and   \Delta BMD  ,

(i)    AM\ =\ BM          (Since M is the mid-point)

(ii)    \angle CMA\ =\ \angle DMB             (Vertically opposite angles are equal)

(iii)    CM\ =\ DM           (Given)

Thus by SAS congruency, we can conclude that :

   \small \Delta AMC\cong \Delta BMD

Posted by

Sanket Gandhi

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