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 figure).

Show that:

(i) ∆AMC ≅ ∆BMD

(ii) ∠DBC is a right angle

(iii) ∆DBC ≅ ∆ACB

(iv) CM = 1/2AB

Answer 1

(i) In ∆AMC and ∆BMD

DM = CM (given)

∠3 = ∠4

(vertically opposite angles)

AM = MB

(as M is the mid-point of AB)

=> ∆AMC ≅ ∆BMD

(by SAS congruency rule)

(ii) ∵ ∆AMC ≅ ∆BMD

=> ∠1 = ∠2 (by c.p.c.t)

But they are alternate angles,

so AC || BD

=> ∠ACB + ∠DBC = 180

(sum of the interior angles on the same side of a transversal is equal to 180°)

But ∠ACB = 90° (given)

=> ∠DBC = 180° = 90°

(iii) In ∆DBC and ∆ACB

∵ ∠DBC = ∠ACB = 90°

BC = BC (common side)

BD = CA

(∵ ∆AMC ≅ ∆BMD)

∴ ∆DBC ≅ ∆ACB

(by SAS congruency rule)

(iv) ∵ ∆DBC ≅ ∆ACB

=> DC = AB

But M mid-point of DC

=> 2CM = CD

=> 2CM = AB [∵ DC = AB]

=> CM = 1/2 AB.

Hence proved.