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. Show that :

(i) ∆AMC ≅ ∆BMD 

(ii) ∠DBC is a right angle. 

(iii) ∆DBC ≅ ∆ACB 

(iv) CM = \(\frac{1}{2}\)AB.

Answer 1

Data : In a 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. 

To Prove: 

(i) ∆AMC ≅ ∆BMD 

(ii) ∠DBC is a right angle. 

(iii) ∆DBC ≅ ∆ACB 

(iv) CM = \(\frac{1}{2}\)AB. 

Proof: (i) In ∆AMC and ∆BMD, 

BM = AM (∵ M is the mid-point of AB) 

DM = CM (Data) 

∠BMD = ∠AMC 

(Vertically opposite angles) 

Now, Side, angle, Side postulate. 

∴ ∆AMC ≅ ∆BMD

(ii) ∆AMC ≅ ∆BMD (Proved) 

Adding AMBC to both sides, 

∆BMD + ∆MBC = ∆AMC + ∆MBC 

∆DBC ≅ ∆ACB. 

∴ AB Hypotenuse = DC Hypotenuse 

∠ACB = ∠DBC = 90° 

∴ ∠DBC is a right angle. 

(iii) In ∆DBC and ∆ACB, 

DB = AC 

∵∆DMB ≅ DMC (Proved) 

∠DBC = ∠ACB (Proved) 

BC is Common. 

Side, angle, side postulate. 

∴∠DBC ≅ ∆ACB. 

(iv) ∆DBC ≅ ∆ACB (proved) 

Hypotenuse AB = Hypotenuse DC 

\(\frac{1}{2}\)AB = \(\frac{1}{2}\)DC 

\(\frac{1}{2}\)AB = CM 

∴ CM = \(\frac{1}{2}\)AB.