Find area of triangle APC in terms of x, y, and z

The triangle ABC is an isosceles right triangle, with the angle at B being the right angle. Let E and D be points between B and C, and between B and A, respectively. Let BA = x, BD = y, BE = z, and P be the point of intersection of segments AE and CD. What is the area of triangle APC in terms of x, y, and z? Express your answer as a fraction with its numerator and denominator completely factorized.

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I have really tried... I even got help from other forums but I do not understand what to do. One procedure is more confusing than the other... I've tried getting the answer using coordinates to no avail... I am beginning to get desperate.

Heres what I've tried:
I tried to get the areas of the three triangles and then Subtract them to get only the area of APC But I got all confused when I tried to subtract.

So the area of ABC: X^2/2
Area of DBC: XY/2
Area of ABE: ZX/2
Area of AEC: (X*√2) (X-2) (sinº45)

The area of AEC I got it using a formula from my book.

This is what I've been told:

Unless I'm missing something, problem 2 is really rather hard. The only way I could do it was to use coordinates. Take B as the origin and BA, BC as the coordinate axes. Then you immediately run into a problem about what to call the two variables, since x and y have already been used in the statement of the question. So you need to use different letters, say s for the variable along the horizontal axis and t along the vertical axis. Find the equations of the lines AE and CD, and solve them to find the (s,t)-coordinates of the point P where they intersect. Then area(APC) = area(ABC) – area(APB) – area(BPC), and the two triangles APB, BPC have base x and heights the two coordinates of P. (There may be a better way to do the question than this messy method, but the answer is sufficiently complicated that I doubt it.)

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as I didn't get it, I asked once more and they gave me this diagram:


And now, I feel really bad cause even with the diagram I don't get the process. To try to get the answer through coordinates does sound like the most sensible thing... However I don't have any numbers so I cannot get the slope...plus the answer has to be expressed in terms of x y and z...and I don't see how you can get a slope from that.

Another Method:

For the second one, let A(.) be the area of the polygon . So A(ABC) = A(AEB) + A(BCD) + A(APC) - A(BPED) (why?) Most of those areas are simple.

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However I cannot get the area of BPED.

In order that the tutors not waste your time posting the same solution methods you've already tried or have already been provided (at all those other tutoring resources), please type out all the information you have so far.

Please provide all of the solutions you have been given elsewhere, along with a clear listing of everything you have tried yourself. Thank you.

Eliz.

Since areaACD = areaABC - areaBCD,
then areaACD = x^2 / 2 - xy / 2 = x(x-y) / 2 : follow that?

BUT you're saying that z MUST be in the solution, correct?

Nice diagram!

The labelling using x, y, z is unfortunate: change them to a, u, v:
so we have AB = BC = a, BD = u, BE = v.
Now we can use the standard y = mx + b to get the equations of AE and CD.

The slope of AE = (a-0) / (0-v) = -a/v; the y-intercept is a:
so equation is y = (-a/v)x + a [1]
Similarly, equation of CD is: y = (-u/a)x + u [2]

Solving [1] and [2] leads to y = au(a - v) / (a^2 - uv):
so that's the y-coordinate of point P, hence the height of triangle CEP;
so area of CEP (since CE = a - v) = y(a - v)

You can tell fairly easily from the diagram that the area
of ACP = ABC - ABE - CEP; so:
ACP = a^2 / 2 - av / 2 - y(a - v) where y = au(a - v) / (a^2 - uv)
So that's the area of ACP in terms of a,u,v.

If you need to show your answer without the "y" appearing, then:
ACP = a^2 / 2 - av / 2 - [au(a - v) / (a^2 - uv)](a - v); simplifies to:
ACP = a^2(a^2 - au - av + uv) / [2(a^2 - uv)]

If you want to really "see" all of the above:
graph ABC with AB = BC = 20, BD=10 and BE = 15;
so you have A(0,20), B(0,0), C(20,0), D(0,10) and E(15,0);
you'll then "see" the coordinates of point P as (12,4).
Substitute a=20, u=10, v=15 in above formula to get area ACP = 40

Thank you so MUCH!

This problem really had me going nuts! Now I am gonna do it again (another time) from scratch!

tyrant_soul said:

Thank you so MUCH!
This problem really had me going nuts! Now I am gonna do it again (another time) from scratch!

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Btw, I answered your PM; you seemed unaware, so I posted here...