Question

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
1.D is the mid point of AC.
2.MD is perpendicular to AC
3.CM=MA 1/2 AB

Answer 1

angle A is common;
angle B=angle M (corresponding angles)
angle C= angle D (BC is parrallel to MD and BC is perpendicular to AC therefore, MD is also perpendicular to AC)..........(i)
ΔAMD≈ΔABC (by AAA)
therefore (AM/AB)=(AD/AC)=1/2                (as ΔAMD≈ΔABC)
AD/(AD+CD)=1/2
2AD=AD+CD
AD=CD
hence d is the mid point of AC
2....  Proved at (i)
3.... In ΔMDA and ΔMDC
  MD is common
  angle MDA=angle  MDC (rt. angle)
  CD=AD  (proved above) 
  therefore ΔMDA is congerant to ΔMDC by RHS
  threfore AM=MC=1/2AB AS M IS THE MID POINT OF AB