Any idea?

What do you mean by projection?Are you taking component?

Please can someone explain me?

Projection means the foot of perpendicular

@above thanks

I am getting the intersection of FE and AK as the circumcenter of triangle AFE.

Length bashing kills the problem, $BK=c^2/a$,$BC=a$,$AL=(ac-c^2)/a$,$AF=b^2/(a+c)$

I think similarity also kills the problem.

tarzanjunior wrote: Length bashing kills the problem, $BK=c^2/a$,$BC=a$,$AL=(ac-c^2)/a$,$AF=b^2/(a+c)$ CAn you show your full solution?

Length basing is much easier, just write all lengths in terms of the sides $a,b,c$

tarzanjunior wrote: Length basing is much easier, just write all lengths in terms of the sides $a,b,c$ Please show your solution(500th post).

Ferid.---. wrote: Let $ABC$ be a right triangle with the hypotenus $BC.$ Let $BE$ be the bisector of the angle $\angle ABC.$ The circumcircle of the triangle $BCE$ cuts the segment $AB$ again at $F.$ Let $K$ be the projection of $A$ on $BC.$ The point $L$ lies on the segment $AB$ such that $BL=BK.$ Prove that $\frac{AL}{AF}=\sqrt{\frac{BK}{BC}}.$ We should use law of sines,$AF \cdot AB=AE \cdot AC$ and angle-bisector theorem.

Ferid.---. wrote: Let $ABC$ be a right triangle with the hypotenus $BC.$ Let $BE$ be the bisector of the angle $\angle ABC.$ The circumcircle of the triangle $BCE$ cuts the segment $AB$ again at $F.$ Let $K$ be the projection of $A$ on $BC.$ The point $L$ lies on the segment $AB$ such that $BL=BK.$ Prove that $\frac{AL}{AF}=\sqrt{\frac{BK}{BC}}.$ $AL = c - BL = c - BK = c - \frac{{{c^2}}}{a} \Leftrightarrow $ $\boxed{AL = \frac{{c(a - c)}}{a}}$ $(1)$ $AF \cdot c = AE \cdot b = \frac{{bc}}{{a + c}} \cdot b \Leftrightarrow $ $\boxed{AF = \frac{{{b^2}}}{{a + c}}}$ $(2)$ $(1), (2)$: $\frac{{AL}}{{AF}} = \frac{{c(a - c)(a + c)}}{{a{b^2}}} = \frac{{c({a^2} - {c^2})}}{{a{b^2}}} = \frac{{c{b^2}}}{{a{b^2}}} = \frac{c}{a} = \sqrt {\frac{{{c^2}}}{{{a^2}}}} = \sqrt {\frac{{{c^2}}}{{{b^2} + {c^2}}}} = \sqrt {1 + \frac{{{c^2}}}{{{b^2}}}} = \sqrt {1 + \frac{{BK}}{{KC}}} = \sqrt {\frac{{BK}}{{BC}}} $

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proposed by Tran Quang Hung, Vietnam