Question

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 the given figure). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = 1/2 AB

Answer 1

(i) In ΔAMC and ΔBMD,

AM = BM (M is the mid-point of AB)

∠AMC = ∠BMD (Vertically opposite angles)

CM = DM (Given)

∴ ΔAMC ≅ ΔBMD (By SAS congruence rule)

∴ AC = BD (By CPCT)

And, ∠ACM = ∠BDM (By CPCT)

(ii) ∠ACM = ∠BDM

However, ∠ACM and ∠BDM are alternate interior angles. Since

alternate angles are equal, It can be said that DB || AC

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

∠DBC + 90º = 180º

∠DBC = 90º

(iii) In ΔDBC and ΔACB, 

DB = AC (Already proved)

∠DBC = ∠ACB (Each 90 )

BC = CB (Common)

∴ ΔDBC ≅ ΔACB (SAS congruence rule)

(iv) ΔDBC ≅ ΔACB 

AB = DC (By CPCT) 

AB = 2 CM

∴ CM =1/2 AB