In the rectangle ABCD, the perpendicular bisector of AC divides the longer side AB in a ratio 2: 1. Then the angle between AC and BD is
option
A. 30 degree
B. 45 degree
C. 60 degree
D. 90 degree

• 90.AC and BD are diagonals of rectangleangle between them is always 90
• 10 years agoHelpfull: Yes(25) No(30)
• If PQ is the perpendicular bisector of AC, intersecting AB at E and AC at 0, then
  AO = OC = AC/2
  AE = 2EB = (2/3)AB
  
  Triangles AOE and ABC are similar.
  => AO/AE = AB/AC
  => (AC/2)/{(2/3)AB} = AB/AC
  => AB = AC(rt3)/2
  
  => Angle CAB = 30
  => Angle between AC and BD is 60 or 120
  
• 10 years agoHelpfull: Yes(8) No(7)
• can somebody put a diagram & prove it
• 10 years agoHelpfull: Yes(4) No(1)
• d
  angle between diagonals=90
• 10 years agoHelpfull: Yes(3) No(1)
• right answer plz
• 9 years agoHelpfull: Yes(3) No(1)
• 1) Imagine that your rectangle has a longer side= 3 (which you can split as 2:1 as you have done on AB).
  
  2) You have stated that perpendicular of Ac cuts AB as 2::1
  
  3) It implies a right angled triangle is formed by jioning said point of cutting, two diagonals (ABCD) intersecting point and point A itself!
  
  4) As 2 is hypotanuse of said triangle other sides are '1' and '(sq rt 3))' (along diagonal)!
  
  5) Your first question is answerd by solving said right angle triangle as ...tan (theta)= 1/(sq rt 3) and theta=30 degrees, which is a well known relation. 30 degrees is the angle in between two sides "sq rt 3 and 2" of said right angle triangle!
  
  Included angle between AC and BD = 30*2= 60 (by a rule of alternate angles by two parallel sides and a traversal)
  
  Answer is 60 degrees!
• 10 years agoHelpfull: Yes(2) No(2)
• Look at the bisector. Let's call the point where it crosses the line AC E and the point where it crosses the line AB F. Then triangle AEF is similar to triangle ACB. (They have the same angles, a right angle, the angle at A and the angles AFE and ACB, which must be equal.
  
  Lets say that the line AC = 2y and AB = 3x, where AF = 2x and FB = x. Then BC = sqrt((2y)^2 - (3x)^2).
  
  The line EF = sqrt((2x)^2 - y^2)
  
  Now use the law of sines. BC/sin(CAB) = AC/sin(B) and EF/sin(CAB) = AF/sin(AEF).
  
  Rearrange both of these equations in terms of sin(B) and sin(AEF). Then, because B and AEF are both right angles, you should be able to set these two equations equal to each other and solve for sin(CAB).
  
  This should enable you to calculate the sin(CAB) and therefore the angle CAB. CAB is identical to DBA, and 180-sum equals angle AEB. .
• 10 years agoHelpfull: Yes(0) No(3)
• what is the right answer???
• 9 years agoHelpfull: Yes(0) No(3)
• angle between diagonals=90
• 9 years agoHelpfull: Yes(0) No(4)
• 60 or 120 ,becuse if it is not then it cannot bisect AB in the ratio 2:1
• 9 years agoHelpfull: Yes(0) No(1)
• If P is the midpoint of AC and its perpendicular bisector intersects AB at Q, then triangles APQ and ABC are similar
  AP/AB = AQ/AC
  (1/2)(x + y)/x = (2x/3)/(x + y)
  1 + y/x = 4/3
  y/x = 1/3
  y/x = 1/3
  Angle between AC and AB = 30
  Angle between AC and BD = 120 or 60
  
  
• 8 years agoHelpfull: Yes(0) No(1)
• AC and BD are diagonals,,so it is 90 degree
• 7 years agoHelpfull: Yes(0) No(1)
• Let P be the mid-point of both AC and BD.
  As we know angles of a rectancle are 90 deg. so since BD is a bisector, angle between BD and DC will remain 45 deg. sim. the angle between AC and CD will be 45 deg. which means DP=CP=45deg. and the three angles of a triangle are 180 deg so angle between AC and BD wiil be 180 -45-45=90 deg.
  Answer=90 degree
• 7 years agoHelpfull: Yes(0) No(1)
• 30degree is the correct answer. don't check other answers in the comment.
• 7 years agoHelpfull: Yes(0) No(2)