euclidean-geometrygeometric transformationgeometric-constructiongeometryplane-geometry
For reference: Calculate the area of triangle $ABC$; if $ED = 16$;
$AB = 10$ and $D = \angle15^o$(Answer:$20$)
My progress:
I didn't get much.
$\triangle ECD – (15^o, 75^o) \implies EC = 4(\sqrt6-\sqrt2), CD = 4(\sqrt 6+\sqrt 2)$
Incenter Th.
$\triangle ABD: \frac{AC}{CI}=\frac{10+BD}{16+EA}\\
S_{CDE} = 4(\sqrt6-\sqrt2)4(\sqrt6+\sqrt2) = 32$
I thought about closing the ABD triangle but as it's any other triangle, I didn't see much of an alternative
$15^\circ$ is not the correct answer. GeoGebra shows the same.
In $\triangle ABD$, traversal $EC$ intersects $BA$ and $BD$ internally and $AD$ externally. Applying Menelaus's theorem,
$ \displaystyle \frac{AC}{CD} \cdot \frac{DN}{NB} \cdot \frac{BE}{EA} = 1 \implies EA = 3 BE$
Or, $EM + MA = 3 BM - 3 EM$.
As $BM = MA$, $BM =2 EM \implies E$ is midpoint of $BM$.
As $CE$ is also the angle bisector, $BC = CM = \sqrt3$ and $CE \perp AB$
If $BE = x$, applying Pythagoras in $\triangle AEC$ and $\triangle BEC$,
$CE^2 = BC^2 - x^2 = AC^2 - (3x)^2 \implies 3 - x^2 = 9 - 9 x^2$
$ \displaystyle x = \frac{\sqrt3}{2}, AB = 2 \sqrt3$
As $BC = CM = MB$, $\triangle BCM$ is equilateral triangle and $\angle ABC = 60^\circ$.
Now given $AD:DC = BA:BC$, $BD$ is angle bisector of $\angle ABC$.
$\therefore \angle ABD = 30^\circ$
(As a side note, $\triangle ACB$ is right triangle with $\angle ACB = 90^\circ$)
Here is a method (generating two different proofs in fact) that does not use Heron's formula.
Having observed the following relationships:
$$20=4^2+2^2, \ \ \ \ 26=5^2+1^2,$$
we can "embed" the triangle into a coordinate grid in this way:
where we have, in the different right triangles, Pythagoras theorem giving:
$$AB=\sqrt{20}=\sqrt{5^2+1^2}, \ \ AC=3 \sqrt{2}=\sqrt{3^2+3^2}, \ \ BC=\sqrt{4^2+2^2}$$
It remains to subtract to the area of rectangle $DEBF$ the areas of the right triangles:
$$4 \times 5 - (5 \times 1/2 + 4 \times 2/2+3\times 3/2)=20-11=9.$$
In fact, with this figure, we do not need to do all these calculations. It is simpler to use Pick's theorem, saying that the area of a polygon whose vertices are on a square grid is given by:
$$A=\color{red}{i}+\frac{b}{2}-1$$
where $i$ is the number of interior points and $b$ the number of boundary points, giving in our case:
$$A=7+\frac{6}{2}-1=9$$
(Many thanks to A-B-C who has recalled me this method I had forgotten).
Best Answer
The key is to recognize that since $\overline{AC}$ bisects $\angle DAB$, the altitudes from $C$ to $\overline{DE}$ and $\overline{AB}$ are congruent. Then since $AB$ is $5/8$ times $DE$, the area of $\triangle ABC$ must be $5/8$ times the area of $\triangle CDE$. You already have the the area of $\triangle CDE$, so the area of $\triangle ABC$ follows.