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 (see Fig. 7.23). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DBC is a right angle.

(iii) ΔDBC ≅ ΔACB

(iv) CM = 1/2 AB

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Answer 1

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Answer 2

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Given,

∠C = 90°, M is the mid-point of AB and DM = CM

(i) In ΔAMC and ΔBMD,

AM = BM (M is the mid-point)

∠CMA = ∠DMB (Vertically opposite angles)

CM = DM (Given)

Therefore, ΔAMC ≅ ΔBMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by CPCT)

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

Now,

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

⇒ 90° + ∠B = 180°

⇒ ∠DBC = 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC (byy CPCT, already proved)

Therefore, ΔDBC ≅ ΔACB by SAS congruence condition.

(iv) DC = AB (ΔDBC ≅ ΔACB)

⇒ DM = CM = AM = BM (M is mid-point)

⇒ DM + CM = AM + BM

⇒ CM + CM = AB

⇒ CM = 1/2AB .