geometric representation

$AB$ 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


  • $\begingroup$ It seems unlikely you'll do $2$ without trigonometry. How would you relate the angle $\beta$ to a sidelength otherwise? I guess $\beta$ could turn out to be some special angle... $\endgroup$ – rschwieb Jun 3 '14 at 13:01
  • $\begingroup$ @ rschwieb - and what about $1)$? $\endgroup$ – bran Jun 3 '14 at 15:53
  • $\begingroup$ That one might be possible :) $\endgroup$ – rschwieb Jun 3 '14 at 18:36
  • $\begingroup$ You are not supposed to ask for answers. You have to show us some work. ASnd this SE is for help, not answers. $\endgroup$ – user122519 Jun 4 '14 at 15:48

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.

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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, $\beta = 60^\circ$.


HINT: Look at chords $AE$, $DC$, and $BF$ and their intersections $G$, $K$ and between $AE$ and $BF$. Since that you already know in $2)$ that the ratio for $\frac{BK}{KF}$ is $2$, which tells you that $BK$ is $2$ times the length of $KF$. Thus, without trigonometric tools, you can find the inter-relationships between the various chords and then find the ratios between the selected inscribed cicles for the selcted triangles inscribed in the initial circle.

  • $\begingroup$ thanks for the hint but you can't take the new info in $2)$ for solving $1)$.. and i disagree with your comment, i don't know how to solve it without trig so what exactly i need to show? i can show my trig solution but i think it's pointless for my question. $\endgroup$ – bran Jun 4 '14 at 21:54
  • $\begingroup$ Well, it depends on if it will benefit the person that is answering your question. $\endgroup$ – user122519 Jun 5 '14 at 4:52

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