Given a triangle ABC, points D, E and F are placed on sides BC, AC and AB, respectively, such that BD : DC = 1 : 1, CE : EA = 1 : 3 and AF : FB = 1 : 4. A line parallel to AB is drawn from D to G on side AC. Lines DG and EF meet at X. If the area of triangle ABC is 120, what is the area of triangle DEX?
I found that AFE is 20, ECD is 20 and BFXD is 75 (i think this is wrong). Subtracting those from 120, I got 5 but it doesn't seem correct. Can anyone please tell me where i went worng?
On
First, observe that $AB:DG=AC:CG=2:1$, so the area of DGC is 30. Notice that $AG:GE:EC=2:1:1$. Thus both DEG and DEC have area 15. Extend AB and DE so they intersect at point Y, then since $DG:AY=GE:AE=1:3, DG:AB=DC:BC=1:2,$ we have $GD=BY$ and thus $AF:FY=GX:XD=2:13$. EGX and DEX share the same height, and $EGX+DEX=DEG$, so DEX has area 13.
I'm not sure how you found those figures you claim to have found.
Notice DEX is CDG-ECD-EGX.
You can see that ABC is similar to CDG with a 1:2 ration of side lengths so that means that the ratio of their areas is 1:4. Thus the area of CDG is $120*\frac{1}{4}=30$. Now, we can find the area of CDE by saying the ratio between CE and AC is 1:4 and by noticing that if we use EC as the base of CDE then its height would be half of what it would be if we used B instead of D. So we can say EDC has area $120*\frac{1}{2}*\frac{1}{4}=15$.
Now to get the area of DEX, we have to subtract out the EGX from the triangle CDG having already subtracted out the area of CDE. We can say that because DG is parallel to AB that the ratio between AG and CG is 1:1. This implies that the ratio between EG and AE is 1:3. Finally, because EGX is similar to AEF, by calculating the area of AFE we can find the area of GXE. The area of AFE can be calculated by noticing the ratio between AF and AB is 1:5 and the ratio of AE and AC 3:4. So the area of AEF is $120*\frac{1}{5}*\frac{3}{4} = 18$ making the area of EGX $18*\frac{1}{9} = 2$.
Thus, we can calculate the area of DEX which is finally $30-15-2 = 13$.