Question

In right triangle ABC, right angles 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 fig.). Show that :

(i) △AMC=△BMD
(II) ∠DBC is a right angle.
(iii) △DBC=△ACB
(iv) CM=1/2 AB​

Answer 1

$\mathfrak{\huge{Solution:}}$

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In △BMB and △DMC ,

We have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

∠DMB = ∠AMC (Vertically opposite angles)

△AMC = △BMD (By SAS)

Hence Proved.

(ii) AC || BD (∠DBM and ∠CAM are alternate angles)

⇒∠DBC + ∠ACB=180° (Sum of co-interior angles)

⇒∠DBC = 90°

Hence proved.

(iii) In △DBC amd △ACB, We have

DB = AC (CPCT)

BC = BC (Common)

∠DBC = ∠ACB (Each=90°) 
△DBC = △ACB (By SAS)

Hence proved.

(iv) AB = CD

⇒1/2 AB = 1/2 CD (CPCT)

Hence, 1/2AB=CM Proved.

 

Answer 2

$\huge\red{Answer}$

In △BMB and △DMC ,

We have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

∠DMB = ∠AMC (Vertically opposite angles)

△AMC = △BMD (By SAS)

Hence Proved.

(ii) AC || BD (∠DBM and ∠CAM are alternate angles)

⇒∠DBC + ∠ACB=180° (Sum of co-interior angles)

⇒∠DBC = 90°

Hence proved.

(iii) In △DBC amd △ACB, We have

DB = AC (CPCT)

BC = BC (Common)

∠DBC = ∠ACB (Each=90°) 

△DBC = △ACB (By SAS)

Hence proved.

(iv) AB = CD

⇒1/2 AB = 1/2 CD (CPCT)

Hence, 1/2AB=CM Proved.

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