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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. 

                  

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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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Sanket Gandhi

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