Geometry homework question - enough data? I've been asked to help with the following school problem on geometry.
In the triangle $\Delta ABC$ one has $AB = 60$, $AC = 80$. Point $O$ is the centre of the circumscribed circle. Point $D$ belongs to the side $AC$. Additionally, one has $AO \perp BD$. One is asked to find $CD$.
(just in case, the answer is $35$)
I am really puzzled, since the information given clearly does not fix the triangle. I know how to solve the problem under the assumption that point $O$ belongs to $BD$. In this case, the solution goes as follows:
Denote $\alpha = \angle OAC$, $\beta = \angle OBC$. 


*

*$\angle ACB = \dfrac{1}{2}\angle AOB = 45^\circ$.

*From the sum of angles of the triangle $\triangle ABC$, one has:
$$\alpha + \beta = 45^\circ$$

*The law of sines for the triangle $\triangle ABC$ gives:
$$\dfrac{AC}{\sin(\beta + 45^\circ)}=\dfrac{AB}{\sin(45^\circ)}$$
From where one can find $\beta$:
$$\beta = \arccos\left( \dfrac{2\sqrt2}{3} \right) + 45^\circ$$

*From the triangle $\triangle AOD$ one finds:
$$CD = AC - AD = AC - \dfrac{AO}{\cos(\alpha)}= AC - \dfrac{AO}{\cos(45^\circ - \beta)}$$


Substituting the value of $\beta$ indeed gives $CD = 35$.
Now, I have two questions:


*

*Is it possible to get the answer without the assumption I have made (or any other one).

*Can anyone present an easier solution? (just in case, this is one of $26$ problems in the $9$th grade quiz in Russian middle school — students are obviously limited in time and are not supposed to use Mathematica and even Stack Exchange)
 A: 
Let $G$ be the point of intersection of $AO$ and $BD.$
$AGB$ is a right triangle.
Drop altitudes $OE$ and $OF$ to sides $AB$ and $AC$ respectively
Since $AOB$ and $AOC$ are isosceles, these altitudes bisect their respective sides.
$AFO$ is similar to $AGB$
$AF:AO = AG: AB$
$AO\cdot AG = 1800$
$AEO$ is similar to $AGD$
$AE:AO = AG: AD$
$AE\cdot AD = 1800$
$AD = 45$
$CD = 35$
A: Draw the line $AO$ and let $E$ be the second point of intersection of $AO$ with the circumcircle of triangle $ABC$ (the first point of intersection being $A$). Then $AE$ is the diameter of the circumcircle and therefore triangle $ABE$ is a right triangle ( $\angle \, ABE = 90^{\circ}$ ). Let $H$ be the itnersection point of $BD$ and $AO$. 

Since by assumption $BD$ is orthogonal to $AO$, and therefore orthogonal to $AE$, segment $BD$ is an the altitude of $ABE$ through $B$. Hence triangles $AHB$ and $ABE$ are similar and thus
$$\frac{AH}{AB} = \frac{AB}{AE}$$ which is equivalent to $$AH \cdot AE = AB^2 = 60^{2}$$
Triangle $AEC$ is right triangle ( $\angle \, ACE = 90^{\circ}$ ) and so is $AHD$, which means they are similar and thus
$$\frac{AD}{AE}=\frac{AH}{AC} $$ which is equivalent to $$AD \cdot AC = AH \cdot AE = 60^2$$ and since $AD = AC - CD = 80 - CD$ and $AC = 80$ get the equation
$$(80-CD)\cdot 80 = 60^2$$ When you solve it you get $CD = 35$.      
A: I don't see how you can solve it without your assumption.
With regards to your method, I believe there is a marginally quicker means:
Note that $\triangle OAB$ is an isosceles right angled triangle $\Rightarrow \angle OAB = \angle ABO = \pi/4$
$AO = BO = 60cos(\pi/4) = 30\sqrt{2}$
Using your definition of $\alpha$, $AD cos\alpha = 30\sqrt{2}$
But then I can't find a way of avoiding a messy means of calculating $cos\alpha$
I think it's just a poor question in the context given.
A: Imho, No.
In coordinates: 
Let $A(0, 0), C(80, 0), B(60 cos(f), 60sin(f) )$
Coordinates of $O$  is $(40, t)$. from $AO = BO$ 
$1600 + t^2 = 3600 - 120*sin(f)*t + t^2$ => $t = \frac{50}{3*sin(f)}$
Equation of $AO$ is $5x - 12*sin(f)*y = 0$, so, equation of BD is 
$12*sin(f)*x + 5*y = 60*12*sin(f)*cos(f) + 60*5*sin(f)$ 
So, coordinates of D is $60*cos(f) + 20$, and $CD = 60(1-cos(f))$
Maybe, I was create same typos and fails, but this method must work :)
