Question

16. In right triangle ABC, right angled at C. M is the mid-point of hypotenuse ABC is joined to M and
produced to a point D such that DM = CM. Point D is joined to point B (see figure) Show that
(1) AAMC ABMD
(ii) 2DBC is a right angle
(iii) ADBC = AACB
(iv) CM 2 AB
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Answer 1

Correct 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

Solution :-

(i) ΔAMC ≅ ΔBMD

Given,

ㅤㅤ∠ACB = 90°

ㅤㅤMid point of AB is M

ㅤㅤBM = AM and DM = CM (M is the mid point)

To prove,

ΔAMC ≅ ΔBMD

Proof,

∠AMC = ∠BMD... [ ∵vertically opposite ang.]

In ∆AMC and ∆BMD,

CM = DM

AM = BM

∠AMC = ∠BMD

∴ ΔAMC ≅ ΔBMD by SAS criteria.

AC = BD and ∠ACM = ∠BDM by CPCT

(ii) ∠DBC is a right angle.

BM = AM (M is mid point)

Then, ∠BDM = ∠MCA (angles opposite to equal sides are equal)

If alternative angles are equal, then sides DB and AC are parallel.

Therefore, their sum would be 180.

∠DBC + ∠BCA = 180°

∠DBC = 180-90

∠DBC = 90°.

Hence, ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

In ∆DBC and ∆ACB,

AB = DC (given)

∠DBC = ∠ACB (perpendicular)

BC = CB (common)

hence, ΔDBC ≅ ΔACB from SAS criteria.

and DC = AB by CPCT.

(iv) CM = 1/2AB

DC = AB (proved above by CPCT)

DC/2 = AB/2

CM = AB/2

CM = 1/2AB

Hence Proved!

Answer 2

Question= In right triangle ABC, right angled at C. M is the mid-point of hypotenuse ABC is joined to M and

produced to a point D such that DM = CM. Point D is joined to point B (see figure) Show that

(1) AAMC ABMD

(ii) 2DBC is a right angle

(iii) ADBC = AACB

(iv) CM 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