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. 7.23). Show that:

(i) Δ AMC ≅Δ BMD

(ii) ∠DBC is a right angle.

(iii) Δ DBC ≅Δ ACB

(iv) CM =AB

It is given in the question that:

∠C = 90^o,

M is the mid-point of AB and DM = CM

(i) In

AM = BM (M is the mid-point)

∠CMA = ∠DMB (Vertically opposite angle)

CM = DM (Given)

Therefore,

By SAS congruence,

(ii) ∠ACM = ∠BDM (By c.p.c.t)

Therefore,

AC parallel to BD as alternate interior angles are equal

Now,

∠ACB + ∠DBC = 180^o (Co-interior angles)

90^o + ∠B = 180^o

∠DBC = 90^o

(iii) In

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC (By c.p.c.t already proved)

Therefore,

By SAS congruence rule,

(iv) DC = AB ()

DM = CM = AM = BM (M is the mid-point)

DM + CM = AM + BM

CM + CM = AB

CM = AB