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

Answer 1

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)

Answer 2

Answer:

0

Step-by-step explanation: