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 = 90o,
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 = 180o (Co-interior angles)
90o + ∠B = 180o
∠DBC = 90o
(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