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) MD AC


(iii) CM = MA =AB


(i) In ,


Given that: M is the mid-point of AB and


MD || BC


Therefore,


D is the mid-point of AC (By converse of mid-point theorem)


(ii) As DM || CB and AC is a transversal line for them


Therefore,


MDC + DCB = 1800 (Co-interior angles)


MDC + 900 = 1800


MDC = 900


MD is perpendicular to AC


(iii) Join MC


In ΔAMD and ΔCMD,


AD = CD (D is the mid-point of side AC)


ADM = CDM (Each 900)


DM = DM (Common)


ΔAMD ΔCMD (By SAS congruence rule)


Therefore,


AM = CM (By CPCT)


However,


AM = AB (M is the mid-point of AB)


Therefore,


CM = AM = AB


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