ABCD is a rectangle, in which BC=2AB. A point E lies on ray CD, such that CE=2BC. Prove that BE perpendicular AC.
let AB = x
then BC = 2x
CE = 2(2x) = 4x
slope BE = 4x/(2x) = 2
slope AC = -x/(2x) , negative since it leans to the left
= -1/2
since 2 is the negative reciprocal of -1/2,
BE and AC are perpendicular.
or
place rectangle ABCD on the x-y grid with B at the origin.
A is (0,x)
B is (,0)
C is (2x, 0)
D is (2x, x)
E is (2x, 4x)
find slope BE and slope AC to show the same result
To prove that BE is perpendicular to AC, we need to show that angle BEC is a right angle.
Given that ABCD is a rectangle, we know that opposite sides are parallel and equal in length. Therefore, AB || CD and BC || AD.
Since BC = 2AB, we can express AB as AB = BC/2.
Given that CE = 2BC, we can express BC as BC = CE/2.
Substituting the values of AB and BC, we have AB = (CE/2)/2 = CE/4.
Since AB = CE/4, we can conclude that triangle ABE is a right-angled triangle, with AB as one of its sides and BE as another side.
Let's assume that angle BEC is not a right angle. In that case, the sum of the angles in triangle BEC should be greater than 180 degrees.
Since the sum of the angles in triangle BEC should be 180 degrees, and angle BEC is not a right angle, this means that the other two angles in the triangle must be acute (less than 90 degrees).
However, in a right-angled triangle, one of the angles is always a right angle (90 degrees). This means that if BEC is not a right angle, then triangle BEC cannot be a right-angled triangle.
But we know from our previous calculations that triangle ABE is a right-angled triangle, with AB as one of its sides and BE as another side.
This contradiction shows that our assumption was incorrect, and angle BEC must be a right angle.
Therefore, BE is perpendicular to AC, as required.
To prove that BE is perpendicular to AC, we need to show that the product of the slopes of AC and BE is equal to -1.
First, let's assign coordinates to the points. Let A be (0, 0), B be (a, 0), C be (a, b), and D be (0, b), where a and b are positive real numbers.
Since BC = 2AB, the length of BC is 2a, and the length of AB is a. Thus, C is (3a, 0).
Similarly, since CE = 2BC, the length of CE is 6a, and the length of BC is 2a. Thus, E is (3a, -2b).
Now, let's calculate the slopes of AC and BE.
The slope of AC is given by the formula: m(AC) = (y2 - y1) / (x2 - x1).
Substituting the coordinates of points A and C into the formula, we get: m(AC) = (b - 0) / (3a - 0) = b / 3a.
The slope of BE is given by the formula: m(BE) = (y2 - y1) / (x2 - x1).
Substituting the coordinates of points B and E into the formula, we get: m(BE) = (-2b - 0) / (3a - a) = -2b / 2a = -b / a.
To prove that BE is perpendicular to AC, we need to show that the product of the slopes of AC and BE is equal to -1. Let's calculate the product:
m(AC) * m(BE) = (b / 3a) * (-b / a) = -b^2 / (3a^2).
We need to show that this product is equal to -1: -b^2 / (3a^2) = -1.
By multiplying both sides of the equation by 3a^2, we get: -b^2 = -3a^2.
Dividing both sides of the equation by -1, we get: b^2 = 3a^2.
Since a and b are positive real numbers, the equation b^2 = 3a^2 holds true.
Therefore, we have proved that BE is perpendicular to AC.