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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.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: \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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