ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that (i) D is the mid-point of AC (ii) MB ⊥ AC (iii) CM = MA = 1/2 AB

(i) In△ABC,Mis the mid point ofABandMD∥BC.Therefore,Dis the mid- point ofAC.i.e.,AD=DC... (1)(ii) SinceMD∥BC, corresponding angles has to be equal∠ADM=∠ACB∠ADM=90∘As∠ACB=90∘is givenSince∠ADMand∠CDMare angles of a linear pairBut∠ADM+∠CDM=180∘Therfore90∘+∠CDM=180∘⇒∠CDM=90∘Thus,∠ADM=∠CMD=90o... (2)⇒MD⊥AC(iii) In trianglesAMDandCMD, we haveAD=CD(from(1))∠ADM=∠CDMandMD=MD(Common)Using SAS criterion of congruence△AMD≅△CMD⇒MA=MC(Since corresponding parts of congruent triangles are equal)Also,MA=12AB, sinceMis the mid-point ofAB.Hence,CM=MA=12AB