Showing three lines pass throw one point (external) ABC is a triangle with acute angles. Take points A′, B′, C′ outside
of the triangle such that:

*

*BA′ = CA′, CB′ = AB′, AC′ = BC′;

*the triangles ABC′, BCA′ and CAB′ do not intersect the triangle ABC;

*these triangles have angles 90◦ at the vertices A′, B′ and C′.
Prove that the lines AA′, BB′, CC′ pass through one point

Im attempting to use Ceva's theorem to solve the question, the approach I took was trying to show AC'F and FC'B are congruent triangles which leads to C'F cutting AB into two equal parts hence being a median and then I could use Ceva's theorem but Im a little stuck in this approach. Are there any suggestions on how I can move forward on this?

 A: Hint: try to use the Ceva's theorem and the following observation: prove that $$
\frac{AF}{FB}=\frac{S_{AC'C}}{S_{BC'C}}=\frac{AC\cdot\sin\angle CAC'}{BC\cdot\sin\angle CBC'}.
$$
A: 
Since Richrow has already provided a proof using Ceva's theorem, I would like to provide a proof using geometric transformation.
Notaion: Instead of $A'$, $B'$ and $C'$, I feel $X$, $Y$ and $Z$ more convenient.
In the figure, we are going to prove that $AX \bot YZ$ and hence $AX$ is an altitude of $\Delta XYZ$
To achieve this, we rotate $YA$ and $YZ$ $90^o$ about $Y$ to obtain $YC$ and $YG$ as shown.
Note that $\angle YCG=\angle ZAY=A+90^o $, $\angle XCY=C+90^o$
Hence $\angle XCG=360^o-(A+90^o)-(C+90^o)=B$
Together with $XC=\frac{BC}{\sqrt 2}$ and $CG=AZ=\frac{AB}{\sqrt 2}$,
we have $\Delta ABC \sim \Delta GCX$
Comparing length ratio, $XG=\frac{AC}{\sqrt 2}=AY$
Note also that by similar triangles, $\angle XGC=A$
and $\angle YCG=90^o+A \implies \angle YCG+\angle YGC=90^o-A$
Thus $\angle AGY+\angle AYG=A+90^o-A+90^o=180^o$
therefore $XG // AY$ and hence $AYGX$ is a parallelogram.
Thus $AX$ is parallel  to $YG$
Since $YG \bot YZ$, therefore $AX \bot YZ$ and $AX$ is an altitude of $\Delta XYZ$
By similar arguments, we can prove that $BY$ and $CZ$ are altitudes of $\Delta XYZ$.
Hence the three lines $AX$, $BY$ and $CZ$ will meet at the orthocentre of $\Delta XYZ$.
