# Prove that $K$ lies on a fixed circle and $AK$ passes through a fixed point.

Consider $$(O, r)$$ being the circumcircle of $$\triangle ABC$$ where $$BC$$ is fixed and $$A$$ is not. $$OE$$ and $$OF$$ are perpendicular to respectively $$AB$$ and $$AC$$ ($$E \in AC$$, $$F \in AB$$). $$(E, EA) \cap (F, FA) = K$$ ($$K \not\equiv A$$). Prove that $$K$$ lies on a fixed circle and $$AK$$ passes through a fixed point where $$\dfrac{BC}{r} = \sqrt 3$$. As can be seen, $$K \in (O, B, C)$$ and $$AK$$ passes through $$O$$. But I can't find out a method to the second one.

The first can be solved like so.

$$\widehat{BOC} = 2 \cdot \sin^{-1}\dfrac{BC}{r} = 120^\circ \implies \widehat{CAB} = 60^\circ$$.

We have that $$\widehat{BKC} = \widehat{BKA} + \widehat{CKA} = \dfrac{\widehat{BEA}}{2} + \dfrac{\widehat{CFA}}{2} = 90^\circ - \widehat{BAE} + 90^\circ - \widehat{CAF}$$

$$180^\circ - 2 \cdot \widehat{CAB} = 180^\circ - 2 \cdot 60^\circ = 60^\circ$$ $$\implies \widehat{BOC} + \widehat{BKC} = 120^\circ + 60^\circ = 180^\circ$$.

That means $$K \in (O, B, C)$$.

• Circle centre $E$, diameter $EA$ Apr 6 '19 at 12:42
• I fixed the problem. Apr 6 '19 at 15:30
• Why is that $\widehat{BOC} = ... = 120^\circ$?
– Mick
Apr 8 '19 at 17:50 I think there is no necessary for given numbers.

first ,to prove $$K$$ is on the fixed circle: $$EO \perp AB \implies AE=BE \implies B \in circle (E,AE)$$,

same reason,$$C \in circle(F,AF)$$ ,connect $$KB,KC,LC,MB$$ $$\angle CKA=\angle CLA, \angle BKA= \angle BMA$$

$$AL$$ is diameter $$\implies \angle CLA= 90^\circ- \angle BAC$$

$$AM$$ is diameter $$\implies \angle BMA= 90^\circ- \angle BAC$$

$$\implies \angle CKA=\angle BKA , \angle BAC= \dfrac{\angle BOC}{2} \implies \angle CKB+\angle BOC=180^\circ \implies K \in circle (O,B,K,C)$$

2nd, to prove AK pass a fixed point:

let $$AK$$ cross $$circle(O,B,K,C)$$ at point $$O'$$, since $$\angle CKA=\angle BKA \implies O'B=O'C$$

$$BO=CO \implies O'=O \implies AK$$ pass $$O$$

3rd to prove $$F \in circle(O,B,K,C)$$ :

$$\angle OFA + \angle BAC =90^\circ \implies \angle OCB=\angle OFA \implies \angle OCB+\angle OFB =180^\circ \implies F \in circle(O,B,K,C)$$

4th to prove $$N \in circle(A,B,C)$$ :

$$\angle BNL + \angle BLN =90^\circ \implies \angle BNL=\angle BAC \implies \angle BNC + \angle BAC=180^\circ \implies N \in circle(A,B,C)$$