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Question -

In right triangle ABC, right angled at C, M isthe mid-point of hypotenuse AB. C is joined to M and produced to a point D suchthat DM = CM. Point D is joined to point B (see figure). Show that
(i) ∆AMC ≅ ∆BMD
(ii) ∠DBC is a right angle
(iii) ∆DBC ≅ ∆ACB



Answer -

Since M is the mid – point of AB.
∴ BM = AM


(i) In ∆AMC and ∆BMD, we have
CM = DM [Given]

(ii) Since ∆AMC ≅ ∆BMD
⇒ ∠MAC = ∠MBD [By C.P.C.T.]
But they form a pair of alternate interior angles.
∴ AC ||DB
Now, BC is a transversal which intersects parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180° [Co-interior angles]
But ∠BCA =90° [∆ABC is right angled at C]
∴ 90° + ∠DBC = 180°
⇒ ∠DBC = 90°

(iii) Again, ∆AMC ≅ ∆BMD [Proved above]
∴ AC =BD [By C.P.C.T.]
Now, in ∆DBC and ∆ACB, we have
BD = CA [Proved above]
∠DBC = ∠ACB [Each 90°]
BC = CB [Common]
∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As∆DBC ≅ ∆ACB
DC = AB [By C.P.C.T.]
But DM = CM [Given]
∴ CM = DC= AB

⇒ CM = AB

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