
$AB$ is a diameter in a circle from point $C$ outside the circle passing to intersect the circle at points $A$ and $B$.
$AC$ intersects the circle at point $F$ and $BC$ intersects the circle at point $E$.
$DC$ is a bisector of $\angle ACB$.
$G$ is the intersecting point of chord $AE$ with $DC$, $K$ is the intersection point of chord $BF$ with $DC$.
$AC=a$, $BC=b$.
$1)$ need to express the ratio between the radius of the Circumscribed circle of $\triangle ADG$ to the radius of the circumscribed circle of $\triangle DKB$, in term of $a$, $b$.
$2)$ given $\measuredangle ACB=\beta$, $\frac{BK}{KF}=2$, need to compute the angle $\beta$.
I solved the question by the law of sine but I'd be glad if one can show how to solve without trigonometric tools
Thanks.
It is quite difficult to solve the problem geometrically completely. At the most is to solve it by as little trigonometric related tools as possible.
To solve this problem, we need two lemmas.
Lemma 1:- The angle bisector theorem. It says “in $\triangle ACB$, if $\angle C$ is bisected by $CD$ (where $D$ is on $AB$), then $AD : BD = AC : CB$. Thus, $AD : DB = a : b$. [For proof, check Wiki.]
Lemma 2:- The length of a chord is given by 2Rsin θ where R is the radius of the circle and θ is the angle at circumference subtending that chord. [For proof, see my answer to problem # 500669.]
Fact:- $\angle AGD = \angle CGE = \angle GEC – \angle ECG = 90^\circ – \frac{β}{2}$. Similarly, $\angle DKB = 90^\circ – \frac{β}{2}$. This further means $\angle AGD = \angle DKB$
$a : b = AD : DB = 2R \sin \angle AGD : 2r \sin \angle BKD = R : r$
The second part:
In $\triangle BCF$, $ \cos \beta = \frac {CF} {CB} = \frac {FK} {KB} = \frac {1}{2}$. Thus, $\beta = 60^\circ$.