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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

Thanks.

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  • $\begingroup$ 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
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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.

enter image description here

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, $\beta = 60^\circ$.

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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.

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  • $\begingroup$ 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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