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 $\dfrac{BC}{r} = \sqrt 3$.

enter image description here

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

  • $\begingroup$ Circle centre $E$, diameter $EA$ $\endgroup$ Apr 6 '19 at 12:42
  • $\begingroup$ I fixed the problem. $\endgroup$ Apr 6 '19 at 15:30
  • $\begingroup$ Why is that $\widehat{BOC} = ... = 120^\circ$? $\endgroup$
    – Mick
    Apr 8 '19 at 17:50

enter image description here

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 $N \in circle(A,B,C)$ :

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.