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: ΔAMC≅ ΔBMD​

Answer 1

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 (ΔDBC ≅ ΔACB)

⇒ DM + CM =AB

[CD=CM+DM]

⇒ CM + CM = AB

[CM= DM (given)]

⇒ 2CM = AB

Hence, CM=1/2AB

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Step-by-step explanation:

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