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 Fig. 7.23). Show that: (i) ΔAMC ≅ ΔBMD (ii) ∠DBC is a right angle.
Solution: It is given that M is the mid-point of the line segment AB, ∠C = 90°, and DM = CM (i) Consider the triangles ΔAMC and ΔBMD: AM = BM (Since M is the mid-point) CM = DM (Given in the question) ∠CMA = ∠DMB (They are vertically opposite angles) So, by SAS congruency criterion, ΔAMC ≅ ΔBMD. (ii) ∠ACM = ∠BDM (by CPCT) ∴ AC || BD as alternate interior angles are equal. Now, ∠ACB +∠DBC = 180° (Since they are co-interiors angles) ⇒ 90° +∠B = 180° ∴ ∠DBC = 90° (iii) In ΔDBC and ΔACB, BC = CB (Common side) ∠ACB = ∠DBC (They are right angles) DB = AC (by CPCT) So, ΔDBC ≅ ΔACB by SAS congruency. (iv) DC = AB (Since ΔDBC ≅ ΔACB) ⇒ DM = CM = AM = BM (Since M the is mid-point) So, DM + CM = BM+AM Hence, CM + CM = AB ⇒ CM = (½) AB
