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.Show that: (i) ΔAMC ≅ ΔBMD (ii) ∠DBC is a right angle. (iii) ΔDBC ≅ ΔACB (iv) CM = 1/2AB Please give me the answer,urgently needed
Answer» Solution:- (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 $\Rightarrow$ ∠DBC + ∠ACB = 180º (Co-interior angles) $\Rightarrow$ ∠DBC + 90º = 180º $\Rightarrow$ ∠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) $\Rightarrow$ AB = 2 CM $\sf{} \because CM =\dfrac{1}{2}AB$ (Figure in attachment)

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) ΔAMC ≅ ΔBMD (ii) ∠DBC is a right angle. (iii) ΔDBC ≅ ΔACB (iv) CM = 1/2AB Please give me the answer,urgently needed

Answer»

Solution:-

(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

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

$\Rightarrow$ ∠DBC + 90º = 180º

$\Rightarrow$ ∠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)

$\Rightarrow$ AB = 2 CM

$\sf{} \because CM =\dfrac{1}{2}AB$

(Figure in attachment)