# 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) $\small \Delta AMC\cong \Delta BMD$

S Sanket Gandhi

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 :

$\small \Delta AMC\cong \Delta BMD$

Exams
Articles
Questions