Prove $EK'\parallel BX\Leftrightarrow EK'=2EK$ in a given triangle $EBC$ with $\angle E=90^\circ\;\ldots$

Question

In a given triangle $EBC$ $\ \angle E=90^\circ$, $D\in BC$ such as $ED\perp BC$, $M'$ is midpoint of $BE$, line $CM'$ intersects the circumcircle of $\triangle EBC$ second time in $X$. $A$ is the second intersection of the line $ED$ with the circumcircle of $\triangle EBC$. $K$ is the intersection of $AX$ with the perpendicular bisector of $BX$. Let $K'\in AX$.

Prove that $XK'=2XK\Leftrightarrow EK'\parallel BX$.

This is the missing intermediate step of my solution to this problem.

My attempt: I see in the case of $EK'\parallel BX$ the triangles $BXM'$ and $M'NE$ are congruent ($N=EK'\cap CX$) thus $M'$ is the midpoint of $XN$ and $\triangle XM'P\sim \triangle XNB$ but I'm failing too show that $B,\,K,\,N$ are collinear.
In the case of $XK'=2XK$ $\ \triangle XKP\sim\triangle XK'B$ hence $BK'\parallel XN$ but I'm missing to show $BK'=XN$.

However, if it's more simple to solve the original problem and hence show the desired result (i.e. "the hole in the solution is as large as the solution itself"), you are free to answer the original question instead.
Thank you.

Answer 1

This is a solution to the original problem:

We have $$\angle BAK = \angle BAX =\angle BCX = \angle BOK$$ so $AOKB$ is cyclic. Since $$\angle ABO =\angle AKO =:\beta \implies \angle ACB =\angle OKT = 90-\beta$$

it is enough to prove $\Delta OKT\sim\Delta CAO$ i.e. $\boxed{{d\over y} = {r\over b}}\;\; (*)$

• Since $BAC$ and $XKT$ are similar we have ${d\over c} = {b\over a}$
• Since $AOM$ and $BKM$ are similar we have ${c\over r} = {x\over e}$
• Since $BMA$ and $KMO$ are similar we have ${a\over y} = {e\over r-x}$

If we multiply these three we get $${da\over ry} = {bx\over a(r-x)}\implies {d\over y} ={rbx\over a^2(r-x)}$$

So $(*)$ will be true iff $$b^2x= a^2(r-x) \iff (a^2+b^2)x = a^2r\iff 4rx=a^2$$

which is true since $ABC$ and $DBA$ are similar.

Answer 2

We use the following diagram. It differs from the one in the question in that the red line $EE'$ is parallel to $BX$ and $K'$ sits in general position on $AX$. We want to show that $XL=2XK$. To do that we show that $BL\perp BX$, which we do by showing that $\triangle E'LA$ is isosceles. Mainly we do a lot of angle and length chasing. Equal angles are shown in the same color (note that red+green=blue). We use the inscribed angle theorem repeatedly: equal chords subtend equal angles and equal angles are subtended by equal chords.

• Because of the right angles at $E$ and $D$, $\angle{BED}=\angle ECB$ (blue).
• By symmetry around KO, $BE'=XE$ and $\angle BEE'=\angle ECX$ (red).
• Since $\angle XCB=\angle E'EA$ (green), $E'A=BX$
• and since $AB=E'X=BE$, $\angle BXA=\angle XAE'=\angle ECB$ (blue).
• Because $E'E \parallel BX$, $\angle BXL=\angle ELX=\angle E'LA$.
• So $\triangle AE'L$ is isosceles and $E'L=E'A=BX=EH$
• By symmetry $\angle LE'B=\angle XEH$ so $\triangle LE'B$ is congruent to $\triangle XEH$ (by SAS) and $\angle BLE'=\angle EHX$ is a right angle.
• Hence $BL$ is perpendicular to $LH$ and $BX$ and $XBLH$ is a rectangle with center $K$.
• So $2XK=XL$ and since $K'$ is constrained to $AX$ , $XK'=2XK \iff K'=L$.