Semi-circle inside of a square

Question

We have a square ABCD, and a semi-circle inside it with AB as its base. We make a tangent line from point C different from CB and mark its point of contact with the semi-circle F. Then we mark the intersection of BD with the semi-circle (different from point B) point E.

What's the area of triangle BEF if AB=10?

Answer 1

It's extremely fast if you just use Pythagoras and similarities. Here's how.

1. Pythagorean Theorem on $\triangle COB$ gives $\overline{CO} = 5\sqrt{5}$.
2. Similarity $\triangle COB \sim \triangle OBH$ yields $\overline{OH} = \sqrt 5$, and $\overline{FB} = 2\overline{BH}= 4\sqrt 5$.
3. Similarity $\triangle OBH \sim \triangle FBP$ gives $\overline{FP} = 4$ and $\overline{PB} = 8$.

Now it is straightforward to calculate the required area by means of successive subtractions from the area of rectangle $PBRQ$:

\begin{eqnarray} \mathcal A_{FBE} &=& \mathcal A_{PBRQ}-\mathcal A_{FBP}-\mathcal A_{EBR}-\mathcal A_{FEQ}=\\ &=&40-16-\frac{25}2-\frac32=10. \end{eqnarray}

Answer 2

I'll skip the calculations because they're probably ugly in a non-informative way.

You know $BC=FC=10$ and $BS=SF=5$. You can then find $CS$ with the Pythagorean theorem. $BCFS$ is a cyclic quadrilateral (since it has opposite right angles), so you can find $BF$ by Ptolemy's theorem.

Now you can work out $m\angle CBF$ and $m\angle BSF$ with the law of cosines. $m\angle EBF=m\angle CNF-45^\circ$ and $m\angle BEF$ is the supplement of half of $m\angle BSF$ because it is an inscribed angle that subtends the same chord as $\angle BSF$.

Now that you have two angles and a non-included side of the desired triangle, you can just bang the rest of it out using the law of sines and the fact that the area of a triangle is half the product of two sides and the sine of the angle between them.

Answer 3

Non-trigonometric solution: Let $S(0,0)$, $B(r,0)$, $C(r,2r)$ and $E(0,r)$. The line $p$ through $BF$ is the polare of the semi-circle from $C$, hence its equation is $$p\colon x\cdot r+y\cdot2r=r^2\iff x=r-2y.$$ From here $F(-\frac35r,\frac45r)$ is easily derived. Call $G(-\frac35,0)$ the foot of the dropped perpendicular from $F$ on $AB$.

Now sum (as usual) the area of the trapezoid $GSEF$ and the triangle $SBE$ and subtract the area of triangle $GBF$ to arrive in $\frac25r^2$.

Answer 4

Sorry for my misunderstanding about posting the solution with external hyperlinks. Since images are acceptable, I divided the solution into 3 different images. Is that acceptable? Or do I need to type it.

Drawing
Solution part 1
Solution part 2

Thanks,
R. de Souza