In right angled 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. Show that: (i) ΔAMC ≅ ΔBMD (ii) ∠DBC is a right angle (iii) ΔDBC ≅ ΔACB (iv) CM = 1/2AB

(i) In△AMCand△BMD, we haveAM=BM∣SinceMis the mid- point ofAB∠AMC=∠BMD∣Vertically opp. anglesandCM=MD∣Given∴By SAS criterion of congruence, we have△AMC≅△BMD(ii) Now,△AMC≅△BMD⇒BD=CAand∠BDM=∠ACM... (1)∣Since corresponding parts of congruent triangles are equalThus, transversalCDcutsCAandBDatCandDrespectively such that the alternate angles∠BDM,∠ACMare equal.∴,BD∥CA.⇒∠CBD+∠BCA=180∘∣Since sum of consecutive interior angles are supplementary⇒∠CBD+90∘=180∘∣Since∠BCA=90∘⇒∠CBD=180∘󔽢∘⇒∠DBC=90∘(iii) Now, in△DBCand△ACB, we haveBD=CA∣From (1)∠DBC=∠ACB∣Since Each=90oBC=BC∣Common∴by SAS criterion of congruence, we have△DBC≅△ACB(iv)CD=AB[Since corresponding parts of congruent triangles are equal]Mis the midpoint ofCD.⇒CM=12CD⇒CM=12AB