I have already solved this problem using trig, however I feel that their must be an easier way to solve this problem using some theorem or property of quadrilaterals that I am forgetting.

geogeometry problem 184

Initially I tried solving this problem using strictly algebra and the angle sum of a triangle theorem. However I quickly found that this just leads me to $x + \angle BDC = 106$. I tried extending the original shape and setting up the problem as a systems of equations and I still couldn't get a singular answer.

Extended image.

When I couldn't find a singular answer with this methods I figured that I must be missing a required theorem relating the angles in a quadrilateral with their diagonals.

I was able to solve for $x$ using the law of sines and cosines, however I feel like this isn't the method that they wanted me to use when solving this problem.

$$\begin{matrix} \frac{DE}{sin(48)} = \frac{AD}{sin(74)} & \frac{AE}{sin(58)} = \frac{AD}{sin(74)} \\ \frac{EC}{sin(30)} = \frac{DE}{sin(44)} & \frac{BE}{sin(16)} = \frac{AE}{sin(58)} \\ \end{matrix}$$

$$BC^2 = BE^2 + CE^2 - 2 \times BE \times CE \times cos(74) $$ $$\frac{sin(74)}{BC} = \frac{sin(x)}{BE} $$ $$x = sin^{-1} \left( \frac{BE \times sin(74)}{BC} \right) $$

So my general question is "What theorem or method should I use to solve this problem?"

Also as I side question, "Can this problem be solved using just the angle sum theorem and algebra?

  • $\begingroup$ x = 30. I constructed the image with geogebra after my initial attempt using just algebra just to verify that there was only one solution and not a family of solutions. $\endgroup$ – Brooks May 16 '16 at 0:27
  • $\begingroup$ It looks to hold in a more general fashion: if $DA=AB$, $\widehat{DAC}=\frac{1}{4}\widehat{DAB}$ and $\widehat{DBC}=30^\circ$, then $\widehat{ACD}=\widehat{DBC}=30^\circ$. Is it just a side-effect of the sine triplication formula? $\endgroup$ – Jack D'Aurizio May 16 '16 at 3:32

Here is a solution without trigonometry:

enter image description here

  1. Draw $BE$ where $\widehat{EBC}=60^{\circ}$
  2. $\widehat{EBC}=2\widehat{EAC}=60^{\circ}$ and $AB=AC$, then $A,E,C$ must lie on a circular arc where $B$ is the center.
  3. $BEC$ is an equilateral triangle.
  4. $\widehat{CED}=2\widehat{CBD}=32^{\circ}$ and $EB=EC$, then $B,C,E,D$ are on a circle
  5. $\widehat{CEB}=2\widehat{BDC}=2x=60^{\circ}$, then $x=30^{\circ}$



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