Last Activity: 4 Years ago
(i) Consider triangles AMC and DMB, We have AM = BM CM = DM [Given] [Given] and LAMC = LBMD [Vertically opposite angles] A AMC ::4 BMD [SAS rule] (ii) As L BAC = L DBA [CPCT, from part (i)] and AB is the transversal. 74 MATHEMATICS-IX So, DB 11 AC ...(i) We have So, AC .l BC DB .lBC ...(ii) [Given] [From (i ) and (ii)] i.e., L DBC is a right angle, i.e., L DBC = 90°. (Ui) Consider triangles ABC and DCB. We have AC = DB [Since l\AMC :: LillMD; result (i)] BC = CB [Common] and L ACB = L DBC .. .'.\ DB ::.'.\ ACB (iu) AB DC .= AB => 2CM = AB \ 1 => CM = -AB. 2 [Each 90°] [SAS rule] [CPCT from part (iii)] [•.• M is mid-point of DC]
Prepraring for the competition made easy just by live online class.
Full Live Access
Study Material
Live Doubts Solving
Daily Class Assignments
Get your questions answered by the expert for free
Last Activity: 2 Years ago
Last Activity: 3 Years ago