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

In Congruent Triangles corresponding parts

are always equal and we write it in short CPCT i e, corresponding parts of Congruent

Triangles.

 

It is necessary to write a correspondence

of vertices correctly for writing the congruence of triangles in symbolic form.

 

Criteria for congruence of triangles:

There are 4 criteria for congruence of

triangles.

SAS( side angle side):

Two Triangles are congruent if two sides

and the included angle of a triangle are equal to the two sides and included

angle of the the other triangle.

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Given: 

In

right angled ∆ABC,

∠C = 90°,

M is the

mid-point of AB i.e, AM=MB & DM = CM.

 

To Prove: 

i) ΔAMC ≅ ΔBMD

ii) ∠DBC is a right angle.

 

Proof:

(i)   In ΔAMC & ΔBMD,

AM = BM                                  (M is the mid-point)

∠CMA = ∠DMB                           (Vertically opposite angles)

CM = DM                                           (Given)

Hence, ΔAMC ≅ ΔBMD

( by SAS

congruence rule)

ii) since, ΔAMC ≅ ΔBMD

AC=DB.

(by CPCT)

∠ACM = ∠BDM (by CPCT)

Hence, AC || BD as alternate interior angles are equal.

Then,

∠ACB + ∠DBC = 180°              (co-interiors angles)

⇒

90° + ∠B = 180°

⇒ ∠DBC = 90°

Hence, ∠DBC = 90°

 

(ii)  In ΔDBC &  ΔACB,

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC ( proved in part ii)

Hence, ΔDBC ≅ ΔACB (by

SAS congruence rule)

(iii)  DC = AB                                              ΔACB)

⇒ DM + CM =AB

[CD=CM+DM]

⇒ CM + CM

= AB

[CM= DM

(given)]

⇒ 2CM = AB

Hence,

CM=1/2AB

 

=ACB

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