In a right triangle abc right angled at c

Asked by admin @ 13/08/2021 in Math viewed by 320 People

"Question 8 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 (see the given figure). Show that: (i) ΔAMC ≅ ΔBMD (ii) ∠DBC is a right angle. (iii) ΔDBC ≅ ΔACB (iv) CM = AB

Class 9 - Math - Triangles Page 120"

Answered by admin @ 13/08/2021

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.

----------------------------------------------------------------------------------------------------

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 (ΔDBC ≅ ΔACB)

⇒ DM + CM =AB

[CD=CM+DM]

⇒ CM + CM
= AB

[CM= DM
(given)]
⇒ 2CM = AB

Hence,
CM=1/2AB

=========================================================

In a right triangle abc right angled at c

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