Question

In right triangle ABC, right angle at C, M is the midpoint of the 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
(i) ∆BMD ≅∆ AMC
(ii) ∠DBM = ∠CAM
(iii) ∠DBC = ∠ACB​

Answer 1

(i) Considering two triangles ΔAMC and ΔBMD:

AM = BM { As given }………………………………………. (i)

∠ CMA = ∠ DMB { Vertically opposite angles }…. (ii)

CM = DM { As given also}……………………………….. (iii)

From above three equation both the triangle satisfies “SAS” congruency criterion

So, ΔAMC ≅ ΔBMD

(ii) From ΔAMC and ΔBMD

∠ ACD = ∠ BDC

Also alternate interior angles of two parallel lines AC and DB.

The sum of two co-interiors angles results to 180°.

So, ∠ ACB + ∠ DBC = 180°

∠ DBC = 180° – ∠ ACB

∠ DBC = 90° { ∠ACB =90° }

(iii) From ΔDBC and ΔACB,

BC { Common side of both the triangle }……. (i)

∠ ACB = ∠ DBC { As both are right angles }….(ii)

DB = AC (by CPCT)………………………………….. (iii)

The triangle satisfies SAS congruency criterion

So, ΔDBC ≅ ΔACB

Learn more about triangles https://brainly.com/question/2773823

#SPJ9