Please Help Me With This ... 10 points reward?

Question

P is a points on the side AB produced of the parallelogram ABCD. Show that the difference in area between the triangles PCB and PAD is constant.

Please help me ... 10 points ... thanks y'all

Answer 1

OK let AB = x
Let AD = y
those are the two different sides of the parallelogram
Let θ = angle DAB = angle CBP

Let z = extended length BP

Recall P is on AB PRODUCED meaning point P is OUTSIDE of the parallelogram.

Triangle PCB has sides y, z and angle θ in between them
Area of PCB = 1/2 * y * z * sin(θ)

Triangle PAD has sides y, (x+z) and angle θ between them
Area of PAD = 1/2 * y *(z+x) * sin(θ)

Difference = 1/2 * x * y * sin(θ)
which is independent of the extended length, z.
Thus the difference is constant for a given parallelogram.

**EDIT**
There is nothing wrong with the wording of the question. Certain people don't seem to understand what "produced" means. It means that AB is extended and hence the point P is outside of the parallelogram, along AB produced or AB extended.

Answer 2

The difference is NOT constant. The sum is constant.
Notice that the area of triangle PDC is constant regardless of the location of point P. (Base and height of this triangle remain the same regardless of P). Therefore the sum of the other two triangles is also constant.

Next, consider a rectangle (which is a parallelogram.) If P is coincident with A then the area of triangle PAD = 0 and the difference in question is half the area of the rectangle. If P is in the midpoint of the side AB then the two triangles are congruent and the difference is = 0.

Extending sides is not applicable here. P would not be "on the side AB" if it were outside the parallelogram. If that is what you meant then the question is not worded correctly. (In fact, its not worded very well as it is.)

Answer 3

You are trying to compare the difference in the area between two triangles. The question asks to prove that as P moves along the side AB that the difference in the area between the two triangles remains constant.

However, the difference in the area between the two triangles is not constant. As one triangle gets smaller, the other gets larger. When P is in the middle of the side AB the two triangles are the same size, so the difference between the two areas is zero. As P moves toward the end of side AB the two triangles become different sizes. Thus there is a difference between the two areas.

In fact, as P approaches either point A or point B the area of one triangle approaches zero while the area of the other triangle approaches one half the area of the parallelogram.

Answer 4

The difference is not constant because if you move P to where it is almost right on top of point A, triangle PAD will have almost 0 area and the other triangle will half nearly half the area of the entire parallelogram. Alternatively, if you put point P directly in the center of segment AB then the two triangle will have equal area, so their difference will be equal to 0. Did you mean to say that the sum of the areas would be a constant--because that would be true?

Answer 5

Hi,

Try this,

Extend CD so that it is
Construct a line segment from P to a point R on Side (CD).

Area of PCB = 1/2 *(PR)*(PB)

Area of PAD = 1/2* (PR)*(PA)

Difference = 1/2 *(PR)*(PB) - 1/2* (PR)*(PA) = 1/2 * (PR) {PB - PA}

Since 1/2 (PR) is constant,

Difference ~ (PB) - (PA)

AB = PA + PB

PB = AB - PA

AB = PA + AB - PA = AB = constant
qed

Hope that helps,
Matt