circleseuclidean-geometrygeometrysolution-verificationtrigonometry
I saw this mathematical puzzle on an Instagram post today. As title suggests, we have a right angled triangle inscribed in a circle with radius $D$ and some angles. The goal is to find the length of $CD$. I'll share my approach as an answer below. I'm not quite sure if my answer is correct, so please feel free to point out any faults in my approach and/or post your own approaches too!
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 $\alpha = 10^{\circ}$.
This is my own approach. I'll add a brief explanation too:
This is the procedure I followed:
1.) Label the triangle $ABC$ and mark all the appropriate angles and side lengths. Notice that since $\triangle ABC$ is obtuse, its circumcenter must lie outside of the triangle itself. Locate circumcenter of $\triangle ABC$ outside and call it point $E$. Connect all the vertices of $\triangle ABC$ to $E$ via $AE$, $BE$ and $CE$. Notice that this means $AB=BE=AE=CE=CD$ because $\triangle ABE$ is equilateral.
(for further explanation of above, point $E$ is a circumcenter of $\triangle ABC$, in that case, $\angle AEB$ must be twice the measure of $\angle ACB$ (inscribed angle theorem), which means it'll be $60$. Because $\angle AEB=60$ and know that $AE=BE$, it follows that $\triangle ABE$ is equilateral)
2.) Notice also that $\angle EAC=\angle ECA=6$. Connect point $D$ with $E$ via segment $DE$. Notice that because segment $CD=CE$ and $\angle ECD=36$, this implies that $\angle EDC=\angle DEC=72$. However, notice that $\angle EBD=96-60=36$, therefore, $\angle BED$ is also $36$. But this implies that segment $BD=DE$
3.) Lastly, notice that this means $\triangle ABD$ is congruent to $\triangle AED$ via the SAS property. This means $\angle BAD=\angle EAD=x$. This means that $2x=60$, therefore, $x=30$.
Best Answer
This is going to be my approach to the problem:
Here's my approach for the problem:
1.) Extend $CD$ to meet $E$ that lies on the circumference. Join the center of the circle $O$ with $E$ via segment $OE$. Since $OE$ is radius, we know that $OE=9$. Notice that $\angle ECB$ is the inscribed angle of $\angle EOC$, therefore we can conclude that $\angle EOB=2\alpha$
2.) Notice that $\angle EOA=\angle EDB=180-2\alpha$, therefore $OE=ED=9$. Now we can use the intersecting chords theorem (which can easily be proven via similarity of triangles in any general cyclic quadrilateral) and conclude that:
$$9\cdot CD=6\cdot12=72$$
$$\Rightarrow CD=8$$