Ptolemy's Riddle III

According to Inscribed Angles Theorem , $\angle DAC = \angle DBC = 60^\circ$ and $\angle ABD = \angle ACD = 60^\circ$ because the each angle of the equilateral triangle is $60^\circ$ .

Therefore, the diagonal $BD$ bisects $\angle ABC$ . Then suppose $AB = a$ , $BC= b$ , $BD = d$ , $AE = m$ , $EC = n$ , and $AC = c = m+n$ so according to Angle Bisector Theorem , $a:b = m:n$ .

In other words, $a\cdot n = b\cdot m$ .

Multiplying both sides by $b-a$ , we will get: $a\cdot n(b-a) = b\cdot m(b-a)$ .

Hence, $a^{2}n + b^{2}m = ab(m+n)$ .

According to Ptolemy's Theorem , $ac + bc = cd$ . Thus, $d = a+b$ , or $BD = AB + BC$ .

Now considering $\triangle ABC$ , according to Stewart's Theorem , we can apply the variables, where $BE = x$ as followed:

$a^{2}n + b^{2}m = (m+n)(x^2 + mn)$

According to Two Secants Theorem , $mn = x(d-x) = x(a+b-x)$ .

Hence, $a^{2}n + b^{2}m = (m+n)x(x + (a+b-x)) = x(m+n)(a+b)$ .

Plugging in the first finding in the earlier equation: $a^{2}n + b^{2}m = ab(m+n) = x(m+n)(a+b)$

Thus, $x = \dfrac{ab}{a+b}$ , or more specifically, $\dfrac{1}{x} = \dfrac{a+b}{ab} = \dfrac{1}{a}+\dfrac{1}{b}$ .

Now substituting $a=3$ & $b=6$ , we will get:

$\dfrac{1}{x} = \dfrac{1}{3}+\dfrac{1}{6} = \dfrac{1}{2}$ . Thus, $x = BE = 2$ .

Finally, $ED = 3+6-2 = \boxed{7}$ .

Since quadrilateral ABCD is cyclic, $\angle ABC = (180 - \angle ADC)^\circ = (180 - 60)^\circ = 120^\circ$ . Also, $\angle ABD = \angle ACD \implies \angle ABD = \angle CBD = 60^\circ$ . Using the Law of Cosines , we see that $AC^2 = AB^2 + BC^2 - (2 \cdot AB \cdot BC \cdot \cos(\angle ABC)) \\ AC^2 = 3^2 + 6^2 - (2 \cdot 3 \cdot 6 \cdot \cos(120^\circ)) \\ AC^2 = 9 + 36 - (36 \cdot -\frac{1}{2}) \\ AC^2 = 45+18 = 63 \\ AC = \sqrt{63} = 3 \cdot \sqrt{7}$ To find the length of $ED$ , we can use the Power of points Theorem , which, in our case, states that $BE \cdot ED = AE \cdot EC$ . First, let's find $AE \cdot EC$ . Seeing that $BD$ bisects $\angle ABC$ , by the Angle Bisector Theorem, we know that $AE = \frac{3}{9} \cdot AC = \frac{1}{3} \cdot 3 \cdot \sqrt{7} = \sqrt{7}$ and $EC = \frac{6}{9} \cdot AC = \frac{2}{3} \cdot 3 \cdot \sqrt{7} = 2 \cdot \sqrt{7}$ Therefore, $AE \cdot EC = \sqrt{7} \cdot 2 \cdot \sqrt{7} = 7 \cdot 2 = 14\\$

Now, to find the length of BE, we can use the Law of Cosines to get $AB^2 = BC^2 + AC^2 - (2 \cdot AC \cdot BC \cdot \cos(\angle ACB)) \\ AB^2 - BC^2 - AC^2 = -(2 \cdot AC \cdot BC \cdot \cos(\angle ACB)) \\ \frac{3^2 - 6^2 - 63}{-2 \cdot 3 \cdot \sqrt{7} \cdot 6} = \cos(\angle ACB) \\ \cos(\angle ACB) = \frac{-90}{-36 \cdot \sqrt{7}} = \frac{5}{2 \cdot \sqrt{7}}$ Using this information, we can use the Law of Cosines again to find $BE$ $BE^2 = BC^2 + EC^2 - (2 \cdot BC \cdot EC \cdot \cos(\angle ACB)) \\ BE^2 = 6^2 + 4 \cdot 7 - (2 \cdot 6 \cdot 2 \cdot \sqrt{7} \cdot \frac{5}{2 \cdot \sqrt{7}}) \\ BE^2 = 36 +28 - (60) \\ BE = \sqrt{4} = 2$ Plugging these into the original equation from the Powers of Points, we get $2 \cdot ED = 14 \\ ED = \frac{14}{2} = \boxed{7}$

$By~ Ptolemy's ~Theorem~BD*AC=3*CD+6*AD,~\implies~BD=9.~~~\because~AC=CD=AD.\\ Using ~Cos~Rule~ in~\Delta~ABC, ~with~\angle~CBA=180-\angle~ADC~of~cyclic~ABCD=120,\\ \therefore~AC=3\sqrt7.\\ \angle s~by ~DC,~~\angle~CBD=\angle~CAD=60.~~\implies~EBA=120-60=60, ~and~ BE~an~angle ~bisector.\\ \therefore~BE=\sqrt{BA*BC \Big (1-~\dfrac{AC^2} {(AB+BC)^2} \Big )} =\sqrt{3*6 \Big (1-\dfrac{63}{(3+6)^2}\Big )}~=~2.\\ ED=BD-BE=9-2=\large~~~\color{#D61F06}{7}.$

$\triangle AED$ and $\triangle BEC$ are similar triangles as $\angle ADB=\angle ACB$ and $\angle AED=\angle BEC$ .

So, $\displaystyle\frac{ED}{EC}=\frac{AD}{BC}\implies EC=ED \cdot\frac{BC}{AD}$

Again, $\triangle ABE$ and $\triangle DCE$ are also similar triangles.

So, $\displaystyle\frac{ED}{EA}=\frac{CD}{AB}\implies EA=ED \cdot\frac{AB}{CD}$

$\displaystyle \implies AC= EA+EC=ED\Big[\frac{BC}{AD}+\frac{AB}{CD}\Big]$

Now, $AC=AD=CD$ as $\triangle ACD$ is equilateral.

$\displaystyle \implies AC=\frac{ED}{AC}(BC+AB) \implies ED=\frac{AC^2}{BC+AB}$

But $AB=3,\, BC=6,\, \angle ABC=180^\circ-\angle ADC=120^\circ$

$\implies AC^2=3^2+6^2-2\times 3\times 6 \cos 120^\circ=9+36+18=63$

So, $ED=\displaystyle\frac{63}{6+3}=\boxed{7}$

By Ptolemy's Theorem, $BD \times AC=3 \times CD+ 6 \times AD \implies BD=9$ .

$\angle ABD = \angle ACD = \angle DAC = \angle DBC = 60^{\circ}$ .

By cosine rule, $AD^2=3^2+9^2-2 \times 3 \times 9 \times \cos(60^{\circ}) \implies AD= 3 \sqrt7$ .

$ar(\Delta ABC)=\dfrac 1 2 \times 3 \times 6 \times \sin(120) = 9 \times \dfrac{\sqrt3}{2}$ and $ar(\Delta DBC)= 63 \times \dfrac{\sqrt3}{4}$ .

$\dfrac{ar(\Delta DAC)}{ar(\Delta ABC)}=\dfrac{DE}{EB}=\dfrac 7 2$ and $DE+EB=9$ .

So, $ED=7$ .

By theorem ,

$\angle ABC + \angle ADC=180$

$\angle ABC + 60=180$

$\angle ABC =120$

Apply cosine law at $\triangle ABC$ :

$AC^2 =3^2+6^2-2(3)(6)(\cos~120)$

$AC^2 =9+36-36(-0.5)$

$AC^2=63$

$AC=\sqrt{63}$

$AC=3\sqrt{7}$

By ptolemys theorem

$3(3\sqrt{7})+6(3\sqrt{7})=3\sqrt{7}BD$

$BD=9$

By intersecting chords theorem

$(BE)(DE)=(AE)(CE)$

However, $CE=3\sqrt{7}-AE$ and $DE=9-BE$ .

Substitute

$(9-BE)(BE)=(3\sqrt{7}-AE)(AE)$

$9BE-BE^2=3\sqrt{7}AE-AE^2$ $\color{#D61F06}(1)$

Apply cosine law at $\triangle ABC$ :

$6^2=3^2+(3\sqrt{7})^2-2(3)(3\sqrt{7})(cos~\theta)$

$\cos \theta=\dfrac{2}{\sqrt{7}}$

Apply cosine law at $\triangle ABE$ :

$BE^2=3^2+AE^2-2(3)(AE)(\dfrac{2}{\sqrt{7}})$

$BE^2=9+AE^2-\dfrac{12\sqrt{7}AE}{7}$ $\color{#D61F06}(2)$

Adding $\color{#D61F06}(1)$ and $\color{#D61F06}(2)$ , we obtain

$BE=\dfrac{\sqrt{7}AE}{7}+1$ $\color{#D61F06}(3)$

Substituting $\color{#D61F06}(3)$ in $\color{#D61F06}(2)$ , we obtain

$\dfrac{6}{7}AE^2-2\sqrt{7}AE+8=0$ $\color{#D61F06}(4)$

Using the quadratic formula in $\color{#D61F06}(4)$ , we obtain

$AE=\sqrt{7}$

Substituting $\sqrt{7}$ for $AE$ in $\color{#D61F06}(3)$ , we get

$BE=2$

Finally,

$ED=9-2=$ $\color{#3D99F6}\large\boxed{7}$

Built a spreadsheet with two parameters -- the radius of the circle and the angle for B. A, C, and D are well defined from the radius. Modified the radius and the angle numbers until the distance from A to B was 3 and the distance from B to C was 6. Took just a couple of minutes. Then the ratio of the difference between the y component of (A - D) / (B - D) times the length of (B - D) is the answer. 7.

By Ptolemy's theorem $BE+ED=9$ and by the intersecting chords theorem $BE\cdot ED=AE\cdot EC$ . By the cosine theorem $AC=\sqrt{63}$ and by the bisector theorem $AE=\frac{1}{3}\sqrt{63},EC=\frac{2}{3}\sqrt{63}$ . It follows that $BE,ED$ fulfill $BE+ED=9, BE\cdot ED=\frac{2}{9}63=14$ , so $BE=2$ and $ED=7$ .

Let $\overline{AE}=t$ and $AB=BC=AC=s$ . Observe that $\triangle AED$ similiar with $\triangle BEC$ . Then $\frac{AE}{BE} = \frac{AD}{BC} \Longleftrightarrow BE = \frac{AE\times BC}{AD} = \frac{ts}{3}$ Observe that $\triangle ABE$ similiar with $DCE$ . Then $\frac{AB}{BE} = \frac{CD}{CE} \Longleftrightarrow \frac{s}{ts/3}= \frac{6}{s-t}$ We have $s=3t$ , $CE=2t$ , and $BE=t^2$ . Because $ABE$ similiar with $DCE$ , we have $\frac{AB}{AE} = \frac{AC}{DE} \Longleftrightarrow DE = \frac{CE \times AE}{BE} = \frac{2t \times t}{t^2}= 2$ We will prove that $BD = AD+DC$ . From Ptolemy's Theorem, $AB \times CD + AD \times BC = AC \times BD \Longleftrightarrow AD+CD=BD$ Then $BD = 3+6=9$ . Therefore, $BE=9-2=\boxed{7}$

By Van Schooten's Theorem, BD = 3 + 6 = 9

By Angle Bisector Theorem, AE : EC = 1 : 2

By Inscribed Angles Theorem, Angle ABC = 120 degrees

By Law of Cosines, AC = 3 sqrt 7

By Stewart's Theorem, BE = 2

Therefore, DE = 9 - 2 = 7.

Thank you po!