8.(ii) 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: (ii) $\small \angle DBC$ is a right angle.

S Sanket Gandhi

In the previous part, we have proved that   $\small \Delta AMC\cong \Delta BMD$.

By c.p.c.t. we can say that    :            $\angle ACM\ =\ \angle BDM$

This implies side AC is parallel to BD.

Thus we can write :            $\angle ACB\ +\ \angle DBC\ =\ 180^{\circ}$          (Co-interior angles)

and,                                            $90^{\circ}\ +\ \angle DBC\ =\ 180^{\circ}$

or                                                                 $\angle DBC\ =\ 90^{\circ}$

Hence $\small \angle DBC$ is a right angle.

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