contest-matheuclidean-geometrygeometrysolution-verificationtrigonometry
As title suggests, in the following figure with some given angles and sides, the goal is to find the measure of length $BD$. This problem originates from Iran from a math olympiad for high school students. I'll post my own approach below as an answer, please let me know if there any faults in it. And please post you own solutions too!
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
I think my solution is quite simple. Since $\triangle BDC$ is isosceles, it follows that $\angle BCD = \angle CBD = 2 \alpha$, and as such, $\angle BCA = 3\alpha$. We then compute that $$ \angle ABD = 180 - \angle CBD - \angle BCA - \angle BAC = 180 - 2\alpha - 3\alpha - 7\alpha = 180 - 12 \alpha. $$ Because also $\triangle ABD$ is isosceles, we decude from this that $$ \angle BAD = \frac{180 - \angle ABD}{2} = 6\alpha. $$ In particular, we find that $\angle CAD = 7\alpha - 6\alpha = \alpha$. This implies that also $\triangle CDA$ is isosceles; whence $|AD| = |DC| = |DB| = |BA|$. It follows that $\triangle BDA$ is even equilateral and as such, its angles are all $60^{\circ}$. Hence $6\alpha = 60^{\circ}$, and thus $\alpha = 10^{\circ}$.
Best Answer
Here's my solution
1.) First we'll notice that $\angle DAC+\angle DCA=120$. We can label $\angle DAC=\beta$ and $\angle DCA=\alpha$. We can "shift" $\triangle ADC$ over segment $AB$ such that the new triangle $\triangle AEB$ is congruent to $\triangle ADC$. Notice that since $\angle BAC=60$, $\angle EAB=\alpha$ and $\angle DAC=\beta$, we can conclude that segment $ED$ is a straight line, or in other words, point $E$, $A$ and $D$ are all collinear since $60+\alpha +\beta=60+120=180$. Note also that $\angle AEB=60$, $AE=3$ and $BE=5$.
2.) Now we know that $ED=8$, we draw a perpendicular from $D$ onto segment $BE$ at point $F$. We can immediately conclude that $\triangle DEF$ is a $30-60-90$ triangle. Therefore $EF=4$ and $DF=4\sqrt3$. We can also tell that $FB=1$. Finally, we can apply the Pythagorean theorem on $\triangle BDF$.
$$BD^2=FD^2+FB^2=48+1=49$$
$$BD=7$$