Given: square ABCD. E is a point on BC; F is a point on CD.

A straight line is drawn from A to E; a straight line is drawn from E to F; a straight line is drawn from A to F.

Hence, four triangles are now produced: ABE, ECF, ADF and AEF.

The area of ABE is known: A1;

The area of ECF is known: A2;

The area of ADF is known: A3.

The problem is to find the area of triangle AEF.

There are many ways that the solution this problem can be approached.

My chosen method was to determine the lengths of AE, EF and AF and thus calculate the area of triangle AEF, using the well-known general formula:

A = √s(s-p)(s-q)(s-r), where the sides of the triangle are p, q and r and s = semi-sum of p, q and r.

To facilitate the necessary algebra, I allocated lengths as follows:

DF = a; CF = b; BE = c.

Hence, CE = a + b -c.

So we have three unknowns, a, b and c and three knowns: A1, A2 and A3.

Thus the three unknowns are determinable. Hence the area of triangle AEF can be found - as is required.

I started by using the 'area of a triangle = half base x height' for each of the three triangles whose areas are known. This produced the three following equations:

2*A1 = (a+b)*c

2*A2 = (a+b-c)*b;

2*A3 = (a+b)*a.

where * signifies multiply.

All of which looks like a promising start. Unfortunately, having performed the usual elimination of two unknowns, thus enabling the third unknown to be determined, I arrived at a hideous cubic equation for that last unknown.

At that point, for me, everything stopped - and I came here!

So: is there a better route to choose in order to attempt to solve this problem?

And if there is, can you show me, please?

Thank you.

Al. (Skywave) / June 16, 2017