Actually vo mujhe 7.1 ka 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 = AB Ans. (i) In AMC and BMD, AM = BM [AB is the mid-point of AB] AMC = BMD [Vertically opposite angles] CM = DM [Given] AMC BMD [By SAS congruency] ACM = BDM ……….(i) CAM = DBM and AC = BD [By C.P.C.T.] (ii) For two lines AC and DB and transversal DC, we have, ACD = BDC [Alternate angles] AC DB Now for parallel lines AC and DB and for transversal BC. DBC = ACB [Alternate angles] ……….(ii) But ABC is a right angled triangle, right angled at C. ACB = ……….(iii) Therefore DBC = [Using eq. (ii) and (iii)] DBC is a right angle. (iii) Now in DBC and ABC, DB = AC [Proved in part (i)] DBC = ACB = [Proved in part (ii)] BC = BC [Common] DBC ACB [By SAS congruency] (iv) Since DBC ACB [Proved above] DC = AB AM + CM = AB CM + CM = AB [ DM = CM] 2CM = AB CM = AB