John's strange quadrilateral

Call $\angle DAC = \theta$ , then to find $\cos\theta$ we can use Law of Cosines:

$15^2 = 9^2 + 21^2 - 2*9*21*\cos\theta$

$\cos\theta = 11/14$

Now using Pythagorean identity, we find $\sin\theta$

$\sin^2\theta + (11/14)^2 = 1$

$\sin\theta = 5\sqrt{3}/14$

We know $\angle CAB = 60^\circ$ because $\triangle ABC$ is equilateral, so $\angle DAB = (60 + \theta)^\circ$ , and we can find $\cos(60 + \theta)$ using the cosine addition formula:

$\cos(60 + \theta) = \cos60\cos\theta - \sin60\sin\theta$

Plugging in our known values gives us:

$\cos(60 + \theta) = -1/7$

So now we use Law of Cosines again on $\angle DAB$ to finally find $BD$ :

$BD^2 = 9^2 + 21^2 - 2*9*21*(-1/7)$

$BD = 24$

Since $AB=BC=21$ and $\angle ABC = 60 ^ \circ$ , $ABC$ must be equilateral and $AC=21$ .

Triangle $ACD$ has sides $AD=9$ , $DC=15$ , and $CA=21$ . Using law of cosines on $\angle CDA$ :

$CA^2=AD^2+DC^2-2 \cdot AD \cdot DC \cdot \cos \angle CDA$

$21^2 = 9^2 + 15^2 - 2 \cdot 9 \cdot 15 \cdot \cos \angle CDA$

$441 = 81 + 225 - 270 \cos \angle CDA$

$135 = -270 \cos \angle CDA$

$\cos \angle CDA = -\frac{1}{2}$

$\angle CDA = 120 ^ \circ$

Since $\angle ABC = 60 ^ \circ$ , $\angle ABC + \angle CDA = 180 ^ \circ$ , so quadrilateral $ABCD$ is cyclic.

Then we can use Ptolemy's:

$AB \cdot CD + BC \cdot AD = AC \cdot BD$

$21 \cdot 15 + 21 \cdot 9 = 21 \cdot BD$

$BD=24$

Since BC=BA, <BCA=<BAC. Since BAC is a triangle, <BCA+<BAC=180-<CBA=180-60=120. Hence, AC=BC=21. Through the cosine rule on triangle ADC, cos <ADC= (AD^2+CD^2-AC^2)/(2 AD DC)=(9^2+15^2-21^2)/(2 9 15)=-1/2. <ADC=cos^-1 (-1/2), and ADC is between 0 to 180 degrees. Hence, <ADC=120, Since <ABC+<ADC=60+120=180, ABCD is a cyclic quadrilateral. We could apply Ptolemy's Theorem on ABCD. AD BC+AB CD=AC BD, so 9 21+21 15=21 BD. Factoring out 21, we get 9+15=BD --> BD=24. The length of BD is 24

Join $AC$ and notice that $\Delta ABC$ is equilateral. Thus $AC=21$ .

By the cosine law, we can find $\cos \angle ACD = \frac{13}{14}$ .

Then $BD^2 = 21^2+15^2-2(21)(15)\cos(60^\circ + \cos^{-1}(\frac{13}{14}))$ .

$BD^2 = 666-630(\cos(60^\circ)(\frac{13}{14}) - \sin(60^\circ)\sin(\cos^{-1}(\frac{13}{14})))$

$BD^2 = 666-630((\frac{13}{28}) - \frac{\sqrt{3}}{2}(\frac{3\sqrt{3}}{14}))$

$BD^2 =576$

$BD =24$

First note triangle ABC. Applying the Law of Cosines, we get $AC^2=21^2+21^2-2 \cdot 21 \cdot 21 \cdot cos60^{\circ} \Rightarrow AC=21$ Next, applying the Law of Cosines on triangle ADC wrt $\angle ADC$ gives us $21^2=15^2+9^2-2 \cdot 15 \cdot 9 \cdot cos\angle ADC \Rightarrow \angle ADC=120^{\circ}$ Thus since the opposite angles of the quadrilateral (in one case $60^{\circ}$ and $120^{\circ}$) are supplementary, the quadrilateral is cyclic. Thus we can apply Ptolmey's Theorem, which states $AC \cdot BD = AB \cdot CD + BC \cdot AD$ $21 \cdot BD = 21 \cdot 15 + 21 \cdot 9 \Rightarrow BD=24$

Clarification: One could have noticed that since triangle ABC was isoceles and the angle created by the two equal sides was $60^{\circ}$, triangle ABC must be equilateral. Proof of "if a quadrialteral's angles are supplementary, the quadrialteral is cyclic" is well known and can easily be found on the internet.

$\bigtriangleup ABC$ is isosceles, since $AB = BC = 21$ , and has an angle of $60^\circ$ , so it is equilateral. This fact tells us that $AC = 21$ . Let $m \angle ADC = \theta$ . Using Law of Cosines on $\bigtriangleup BCD$ , we find that $15^2 + 9^2 - 2 \cdot 9 \cdot 15 \cdot \cos (\theta) = 21^2$ $\displaystyle \to \cos (\theta) = \frac{441 - 225 - 81}{- 2 \cdot 9 \cdot 15} \to$ $\displaystyle \cos(\theta) = -\frac{135}{270} =-\frac{1}{2}$ $\to \theta = 120^\circ$ .

Because $\angle ADC$ and $\angle ABC$ are supplementary, quadrilateral $ABCD$ can be inscribed in a circle, and so we can use Ptolemy's Theorem on it. Ptolemy's theorem states that for any quadrilateral inscribed in a circle, the sum of the products of the lengths of the opposite sides is equal to the product of the lengths of the diagonals. This means that for our quadrilateral, $AB \cdot CD + BC \cdot AD = BD \cdot AC$ . Substituting in the lengths we know, we can solve for $BD$ : $21 \cdot 15 + 21 \cdot 9 = 21 \cdot BD \to$ $21 \cdot (15 + 9) = 21 \cdot BD \to 24 = BD$ . Therefore, $24$ is our answer.

Because $AB=BC$ , then $\angle ACB=\angle BAC=\frac {1}{2}(180 ^ \circ - \angle ABC)=60 ^ \circ.$

$ABC$ is an equilateral triangle. So, $AC=21$ .

By Cosine's Law on $\Delta ACD,$ We get

$\cos \angle ACD=\frac {AC^2+CD^2-AD^2}{2 \cdot AC \cdot AD}$

$\iff \cos \angle ACD=\frac {21^2+15^2-9^2}{2 \cdot 21 \cdot 15}$

$\iff \cos \angle ACD=\frac {13}{14}$

Because $0 ^ \circ \leq \angle ACD \leq 180 ^ \circ$

Then $\sin \angle ACD=\frac {3 \sqrt{3}}{14}$

By Cosine's Law on $\Delta BCD,$ We get

$BD^2=BC^2+CD^2-2 \cdot BC \cdot CD \cdot \cos \angle BCD$

$\iff BD^2=9(7^2+5^2-2 \cdot 7 \cdot 5 \cdot \cos (\angle ACD +60 ^ \circ)$

$\iff BD^2=9(74-70 (\frac {1}{2} \cdot \frac {13}{14}-\frac {\sqrt{3}}{2} \cdot \frac {3 \sqrt{3}}{14}))$

$\iff BD^2=9 \cdot 64$ $\iff BD=24$

Apply cosine rule on triangles $ABC$ and $ADC$ on angles $ABC$ and $ADC$ respectively. We will solve for $AC$ then $\angle ADC$ . $AC^2=21^2+21^2-2(21)^2 \cos 60^\circ=21^2=15^2+9^2-2(15)(9)\cos \angle ADC \Rightarrow \cos \angle ADC=-\frac{1}{2} \Rightarrow \angle ADC=120^\circ$ (since the quadrilateral is convex). Then $\angle ABC+\angle ADC=180^\circ$ . Therefore quadrilateral $ABCD$ is cyclic and by Ptolemy's Theorem we have $BD=24$ .

Since $AB=BC$ , $\angle BAC=\angle ACB=\frac{180^\circ-60^\circ}{2}=60^\circ$ . $ABC$ is an equilateral triangle. $AC=21$ . Applying Cosine rule to triangle $ADC$ , $21^2=9^2+15^2-2(9)(15)(\cos\angle ADC), \cos\angle ADC=-\frac{1}{2}, \angle ADC=120^\circ$ . Since $\angle ABC+\angle ADC=60^\circ+120^\circ=180^\circ$ , $ABCD$ is a cyclic quadrilateral. By Ptolemy Theorem, $(21)(BD)=(21)(15)+(21)(9)$ . $BD=24$ .

Since ABC is isosceles with vertex 60 degrees, then the rest of the angles must be 60, so it's equilateral. Thus, AC=21, by LOC on ADC we get angle ADC=120. Since the opposite angles are thus supplementary, ABCD is cyclic. By Ptolemy, BD AC=BD 21=21(9+15)=21*24, so BD=24.

Since triangle ABC has two side lengths equal to each other, and between the two sides is a 60 degree angle, triangle ABC is equilateral. Thus, $AC=21$ . By the Law of Cosines, $\cos \angle CAD = \frac{9^2+15^2-21^2}{2 \cdot 9 \cdot 15} = -\frac{1}{2}$ .

Thus, $m\angle CAD = 120^{\circ}$ . Observe that $m\angle ABC + m\angle CAD = 180^{\circ},$ so quadrilateral ABCD is cyclic. By Ptolemy's Theorem, $AB \cdot CD + BC \cdot AD = AC \cdot BD.$ Substituting in the known values results in $BD = \boxed{24}.$

Extend CB to E s.t. BE = 35. Note that if we can show AE = 49, by SSS, ABE and ADC are similar, implying ABCD is conyclic (angles ADC and ABC sum to 180 degrees)

Let the foot of perpendicular from A to BC be G

AG = 21(rt3/2)

GE = 35 + 21/2

by pythagoras theorem, AE = 49.

Thus ABCD is conyclic. Now apply Ptolemy's theorem, which gives us:

21(15) + 21(9) = 21(BD)

BD = 24.

AB = BC = 21 and [\angle ABC = 60 ^ \circ] ∴ ABC is an equilateral triangle. ∴ A(triangle ABC) = \frac {441 (\sqrt{3})}{4} by heron's formula A(triangle ACD) = \frac{135 (\sqrt{3})}{2} ∴ a(ABCD) = \frac{711 (\sqrt{3})}{4} but a(ABCD)=a(ADC)+a(BAD) ∴a(ACD)+a(BAD)= \frac{135 (\sqrt{3})}{2}

1] a(ACD)=\frac{\sqrt{(1296-BD²)(BD² - 36)}}{4} 2]a(ABD)=\frac{\sqrt{(900-BD²)(BD² - 144)}}{4}

∴\frac{\sqrt{(1296-BD²)(BD² - 36)}}{4} + \frac{\sqrt{(900-BD²)(BD² - 144)}}{4} = \frac{135*(\sqrt{3})}{2}

by solving we get positive value of BD = 24

Using Law of Cosines on $\Delta ACD$ gives $\angle ADC=120^{\circ}$ ; therefore $ABCD$ is cyclic. By Ptolemy's Theorem, $21\cdot 9+21\cdot 15=21\cdot BD$ ; therefore $BD=\boxed{24}$ .

Triangle ABC is equilateral triangle.Consider triangle ACD. Use cos rule to find angle ADC. Cos ADC = (9^2+15^2-21^2)/(2 9 15), then we find the angle ADC = 120 degrees.By using sin rule : 21/sin120 = 15/sin DAC. Then the angle BAD = 60 + DAC, and BD ^2 = 21^2+9^2 - 2(9)(21) cos BAD = 21^2+9^2 + 54 = 576. So BD = 24.

In isosceles triangle ABC: angle(BAC) = angle(BCA) = 60, so AC = 21. The cosine rule in triangle ACD gives angle(A2) = 38,213 degrees. Angle(A) in triangle ABD is 98,213 degrees. The cosine rule in triangle ABD gives BD = 24.

$AC=21$

Cosine law give: $\angle ADC=\arccos \left( \frac{(CD^2+AD^2-AC^2)}{2 \times CD \times AD} \right)=\arccos \left(-\frac{1}{2} \right)$

So, $\angle ADC=120^{\circ}$

$\angle ABC+ \angle ADC =180^{\circ}$

So, $ABCD$ is cyclic.

By ptolemy theorem

$BC \times AD+ AB \times CD = AC \times BD$

then

$BD=AD+CD=9+15=\fbox{24}$

easy just apply ptolemy's theorm which states sum of product of lengths of opposite sides of a cyclic quadrilateral is equal to the product of the lengths of the diagonals of that quadrilateral.

We join A and C.Since AB=BC,

$\angle BAC=\angle BCA$

But given, $\angle ABC=60$ .So $\angle BAC=\angle BCA=60$ .So $\Delta ABC$ is an equilateral triangle.This implies that AC=21.Using the law of cosines in triangle ACD,we get

$cos(D)=-.5\implies \angle D=120$

$\angle ADC+\angle ABC=180$

From this,we can conclude that quadrilateral ABCD is cyclic.We know by Ptolemy's theorem that

$AC\cdot BD=AB\cdot CD+AD\cdot BC$

Substituting values give us $BD=24$ .

If we draw in $AC$ , we see that $\triangle ABC$ is equilateral. Note that $\triangle CDA$ is a 3-5-7 triangle; thus, $\angle ADC = 120^\circ$ . Since $\angle ABC + \angle ADC = 180$ , we see that $ABCD$ is cyclic. Thus, by Ptolemy's Theorem, $BD \cdot 21 = BD \cdot AC = AB \cdot CD + BC \cdot DA = 24 \cdot 21$ , so $BD = \boxed{24}$ .