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 Fig. 7.23). Show that: 

(i) ∆ AMC ≅ ∆ BMD 

(ii) ∠ DBC is a right angle. 

(iii) ∆ DBC ≅ ∆ ACB (iv) CM = \(\frac{1}{2}\) AB

Answer 1

(i) In ∆AMC and ∆BMD,

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

∠AMC = ∠BMD (Vertically opposite angles)

CM = DM (Given) 

∠∆AMC ∠∆BMD (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 ∆DBC and ∆ACB, 

DB = AC (Already proved) 

∠DBC = ∠ACB (Each 90º) 

BC = CB (Common) 

∠∆DBC ∆ACB (SAS congruence rule) 

(iv) ∆DBC ∆ACB  

∠AB = ∠DC (By CPCT) 

∠AB = 2 CM

∴ CM= \(\frac{1}{2}\) AB