+0 A triangle is partitioned into three triangles and a quadrilateral by drawing two lines from vertices to their opposite sides. The areas of the three triangles are 3, 7, and 7, as shown. What is the area of the shaded quadrilateral?
0 its ok to post questions that one thinks are fun and have a good solution even though i know the answer right? Its like sharing what I think is a good experience.
+399 
Great problem! I had to think some time about this one.
We first notice these are all cevian lines, so mass points work perfectly!
Becuase [BFC] = [FCE] then \(\overline{BF} = \overline{FE}\).
Similarly, \(\frac{\overline{DF}}{\overline{FC}}=\frac{3}{7}\).
Label the mass of D as 7, the mass of C as 3, so the mass of F is 3+7 = 10.
Because \(\overline{BF} = \overline{FE}\), then the mass is split evenly between B and E, so they are both 5.
The mass of D is 7, and the mass of B is 5, so the mass of A is 7-5 = 2.
Becuase mass of A * length of AD = mass of B * length of BD, we get that the ratio of the lengths of AD : BD is \(\frac{AD}{BD}=\frac{5}{2}\).
Similarly, \(\frac{AE}{EC}=\frac{3}{2}\). With these ratios, we also know the ratios of the area of some triangles, which we can use to our advantage.
Set the area of the shaded quadrilateral as x, and the area of the triangle ABC as Y.
We get two systems, by finding the area of triangle ADC and AEB, each two ways.
\(\begin{cases} \frac{3}{5}y=x+3 \\ \frac{5}{7}y=x+7 \end{cases}\). We solve and get x = 18. So the area of the shaded quadrilateral is 18.
0 Amazing solution thankyou! Mass points was a very nice way of going about solving this. Personally i just used simple triangle properties like ratio bases with same altitudes. I think mass points might be easier though.
