8. In right triangk ABC, right angkd at DAC, M is the mUl-point of hypotenuseAB. C isjoined to M and produced toa point D such that DM = CM. PointBCD is joined to point B (see figure ).Show that:(i) 4AMC ::4BMD(ii°) L DBC is a right angk.(iii) ADBC ::AACB(iv) CM = .!_ AB.2

(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]