areacontest-matheuclidean-geometrygeometrytrigonometry
The question is as stated in the title, in the figure given below, find the area of $\triangle BEC$. I must admit this was a challenging problem and the solution that I came up with (which will also be posted as an answer) is pretty complicated and "messy". So, I'd like to see if there are any better approaches that may be simpler as well.
This will be my approach to this problem. I shall add a brief explanation as well:
So this is how I go about it:
Please note that I forgot to mark the point of intersection of the two diagonals, therefore I will be referring to it as point $X$ throughout my explanation.
1.) Locate point $A'$ by extending segment $AC$ such that $A'A$=$AC$ and $\angle A'AD$=75. Connect point $D$ and $A$. Notice that $\triangle A'AD$ and $\triangle ABC$ are congruent via the SAS property.
2.) In $\triangle A'DC$, because the base $A'C$ is divided into two equal segments by a median, we can conclude that $\triangle A'AD$ and $\triangle ADC$ have the same area via a well-known lemma (the proof of which is trivial). By extension, via congruency that we proved earlier, we can also conclude that $\triangle ABC$ and $\triangle ADC$ also have the same area. This information will be helpful, as knowing the area of just one triangle will help is reach our desired result.
3.) Locate point $F$ on segment $A'A$ and connect it to point $D$ such that $\angle DFA$=45, $\angle ADF$=60 and note than $\angle DXA$=45. This implies that $\triangle FDX$ is an isosceles right triangle where segment $DF$=$DX$. Note further that $\triangle FAD$ is congruent to $\triangle XBC$ via the ASA property. Hence, we can conclude that segment $FD$=$DX$=$XB$. This proves that point $X$ is the midpoint of segment $BD$.
4.) Extend segment $DA$ to point $E$ and connect it to point $B$ such that segment $EB$ is perpendicular to segment $AE$. Notice that this forms a right triangle of the type 30-60-90, where $\angle EBD$=60. It is trivially known that, in such a triangle, the side opposite to the angle measure of 30 is half of the largest side hypotenuse (this can be proven via trigonometry as well). However, half of segment $BD$ would be equivalent to $DX$ and $XB$ as point $X$ was found to be a midpoint. Thus we can conclude that segment $DX$=$XB$=$EB$. Connect point $E$ and $B$, because $\angle EBD$=60, $\triangle EXB$ is equilateral, therefore $EX$=$XB$=$EB$.
5.) Above implies that $\angle AEX$=30, and $\angle AXE$=75, therefore $\angle EAX$ must be 75, thus $\triangle AEX$ is isosceles and segment $AE$=$BE$=$BX$=$EX$. This proves that $\triangle AEB$ is also an isosceles right triangle, therefore $\angle ABX$=60-45=15. However, this implies that $\angle ABC$=$\angle ACB$=75, and $\angle BAC$=30. Therefore $\triangle ABC$ is an isosceles triangle as well, thus line segment $AC$=$AB$= 2 units. We drop a perpendicular from point $B$ on segment $AC$ to meet at point $G$. $\triangle ABG$ is a right triangle of type 30-60-90, and as established above, the base opposite to angle measure 30 must be half of the hypotenuse, thus we can conclude that segment $BG$ is 1 unit.
6.) Since we now have a height and a base, we can compute the area of $\triangle ABC$, which is 1 unit^2. However as we established earlier, $\triangle ABC$ and $\triangle ADC$ have equal areas, therefore the area of $\triangle ADC$ is also 1 unit^2 and thus, the total area of the quadrilateral, our answer, is 2 unit^2
Let us denote by $X$ the area to be computed, and let us use letters for the relevant vertices as in the following picture:
Then applying Menelaus in $\Delta EFB$ w.r.t. the line $CDA$ we get: $$ 1= \frac{CE}{CF}\cdot \frac{DF}{DB}\cdot \frac{AB}{AE} = \frac{6+9}{9}\cdot \frac{3}{3+9}\cdot \frac{X+3+6+9}{X+3}\ , $$ which gives a first order equation in $X$ with unique solution $X=54/7$.
Best Answer
The figure is obtained by putting $\Delta BCE$ and $\Delta BAE$ together, with $BA$ overlapping $BC$.
Note that $$\angle CBE + \angle BEA = 90^o \implies \angle E_1BE_2=90^o$$
By Pythagoras theorem, $E_1E_2=\sqrt 8$
Noting that $$E_1A_C^2+E_1E_2^2=8+64=72=E_2A_C^2$$
By Converse of Pythagoras' Theorem, $\angle E_2E_1A_C=90^o$
Thus $\angle BE_1A_C =135^o$
Hence the required area is$$\frac{1}{2}(2)(8)\sin 135^o = \frac{1}{2}(2)(8)\frac{\sqrt 2}{2}=4\sqrt 2$$