Question

In right triangle ABC, right angled at C o
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:
DM = CM. Point D is joined to point B
(see Fig. 7.23). Show that:
(1) Angle AMC congrence angle BMD
(ii) angle DBC is a right angle.
(iii) Angle DBC congrence Angle ACB
Fig. 7.23
(iv) CM = 1/2 AB

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Answer 1

Answer:

4 proofs I given below

Step-by-step explanation:

(1) 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