# Find area of quadrilateral in triangle. [closed]

What is the area of $$HIJK$$ quadrilateral, if the area of $$ABC$$ triangle is $$70$$, $$BE=ED=DA$$, and $$BF=FG= GC$$?

## closed as off-topic by Crostul, verret, Alexander Gruber♦Jan 8 at 22:04

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• I got $\frac{50}{3}.$ – Michael Rozenberg Jan 8 at 19:59
• There should be a solution using Ceva's thm or sth. – user614671 Jan 8 at 20:02
• The area is $9$. Tthe points $B,I,K$ are collinear and the line containing them intersect $AC$ at its midpoint $M$. $\triangle HIK$ is bounded by 3 cevians $AF$, $BM$ and $CD$ with $$(x,y,z) \stackrel{def}{=}\left(\frac{CF}{BF}, \frac{AM}{MC}, \frac{BD}{AD}\right) = (2,1,2)$$ By Routh's theorem, area of $\triangle HIK$ is $$\frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}\cdot 70 = \frac{9}{140}\cdot 70 = \frac92$$ By a similar argument, area of $\triangle IJK$ is also $\frac92$. – achille hui Jan 9 at 14:28

Let's put the area of 70 on hold for a moment. Furthermore, given the listed constraints and the implicit assumption that the problem is solvable using only those constraints, we choose any proportions for the outer triangle we like.

### Let's examine the case of a right isosceles triangle with hypotenuse $$BC$$ and a base length of 3. It has an area of $$\frac{9}{2}$$.

Letting the origin $$(0,0)$$ rest at $$A$$

• $$B$$ is at $$(0,3)$$.
• $$C$$ is at $$(3,0)$$.
• $$D$$ is at $$(0,1)$$.
• $$E$$ is at $$(0,2)$$.
• $$F$$ is at $$(1,2)$$.
• $$G$$ is at $$(2,1)$$.

This gives us the following lines

• $$AF$$ is $$y=2x$$
• $$AG$$ is $$y=\frac{1}{2}x$$
• $$DC$$ is $$y=-\frac{1}{3}x+1$$
• $$EC$$ is $$y=-\frac{2}{3}x+2$$

Considering the four batches of two equations with two unknowns yields the coordinates of the remaining four points.

• $$H$$, from $$y=2x,\;y=-\frac{1}{3}x+1$$ is $$(\frac{3}{7},\frac{6}{7})$$.
• $$I$$, from $$y=2x,\;y=-\frac{2}{3}x+2$$ is $$(\frac{3}{4},\frac{3}{2})$$.
• $$J$$, from $$y=\frac{1}{2}x,\;y=-\frac{2}{3}x+2$$ is $$(\frac{12}{7},\frac{6}{7})$$.
• $$K$$, from $$y=\frac{1}{2}x,\;y=-\frac{1}{3}x+1$$ is $$(\frac{6}{5},\frac{3}{5})$$.

We can express the general form for the area of a quadrilateral based on it's vertices using vector arithmetic $$A=\frac{1}{2}\vert(\vec{J}-\vec{H})\times(\vec{K}-\vec{I})\vert$$ We can plug this into Wolfram Alpha as

abs(Cross[([12/7,6/7]-[3/7,6/7]),([6/5,3/5]-[3/4,3/2])])*.5


### Scaling that for our area 70 triangle, gives a final area of 9.

This differs from Peter Foreman's answer, so one of us must have made a typo or a false assumption. Frankly mine just looks wrong, judging from your diagram. In any case he deserves credit for advocating such a brute-force approach.

• Thanks! it seems I needed to be more specific to wolfram about what I was trying to do. – ShapeOfMatter Jan 9 at 15:04

The area is $$\frac{741}{68}$$ I'm quite sure. You can find the coordinates of $$H,I,J,K$$ by using the equations of each straight line and taking $$A$$ as the origin. I then used Wolfram: Alpha to find the area given the coordinates of each vertex.