In right triangle ABC, right angled at C, M is the mid-point of hypote

1.	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» (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

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»

(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