Prove that $TK=TO$

Question

Given $\triangle ABC$ such that $\angle A=90^\circ$ inscribed in circle with center $O$. Let $D$ be the feet perpendicular from $A$ to $BC$ and $M$ be the mid-point of $BD$. Draw the line $AM$ and let it intersect the circumcircle at $X$. Let $K$ be the point on $AX$ such that $OK//XC$. Lastly, denote $T$ as the intersection of the perpendicular from $AX$ at $K$ to $XC$. Prove that $TK=TO$

I do some angle chasing but I haven't use anything that the problem given like the perp and midpoint for example, as I dun know how I could apply it. BTW, my approach is to prove that $\angle{BCA}=\angle{TOC}$ or perhaps prove that $\triangle ABX$ is similar to $\triangle TOC$. Please help

Answer 1

Our solution relies on removing all the "annoying" points; essentially, $T$ and $K$ do not have many properties that we can use, so we attempt to rid them from our equations.

As you noted, we only need to have $\triangle ABX \sim \triangle TOC$, and then we are done. Since $\angle TCO = \angle BAX$, we only need to prove that $\frac{TC}{XA} = \frac{OC}{AB}$, or that $$2 \cdot TC \cdot AB = 2 \cdot OC \cdot XA = BC \cdot XA = BA \cdot CX + BX \cdot AC$$ by Ptolemy's theorem. Notice that $CX - CT = XT$, so we now want to prove that $$AB \cdot TC = BA \cdot XT + BX \cdot AC$$ or that $AB \cdot CX = 2 \cdot BA \cdot XT + BX \cdot AC$. Now, $\triangle XKT \sim \triangle ABC$, so we have $XT \cdot AB = XK \cdot BC$, and our equation now becomes $$AB \cdot CX = 2 \cdot XK \cdot BC + BX \cdot AC$$ and we have successfully removed the point $T$ from our equation. Now, to remove $K$, we notice that $\triangle OMK \sim \triangle CMX$, so $\frac{KX}{OC} = \frac{MX}{MC}$ and therefore substituting for $XK$ we now want to prove $$AB \cdot CX = BC^2 \frac{MX}{MC} + BX \cdot AC$$ We now make our final change to this equation, and then apply some trigonometry to finish the problem. By Power of a Point, we have $BM \cdot MC = AM \cdot MX$, or that $\frac{MX}{MC} = \frac{BM}{AM} = \frac{MD}{AM} = \sin \angle MAD$. Thus, we have that we want to prove $$AB \cdot CX = BC^2\sin \angle MAD + BX \cdot AC$$

Now, notice that $AB = BC \cos \angle ABC$, $CX = BC \cos \angle BCX$, $BX = BC \sin \angle BCX$, and $AC = BC \sin \angle ABC$, so we wish to prove that $$BC^2(\cos \angle ABC\cos \angle BCX - \sin \angle BCX\sin \angle ABC) = BC^2 \sin \angle MAD$$ However, it is well-known that for any angles $x$ and $y$, $\cos x \cos y - \sin x \sin y = \cos{(x+y)}$, so all we wish to show now is that $$\cos{(\angle ABC + \angle BCX)} = \sin \angle MAD$$ which is true since $\angle ABC + \angle BCX = \angle ABM + \angle BAX = \angle AMD$, so, since all our steps are reversible, we are done.

Answer 2

1. It's quite evident that $\angle BAC=90^\circ$ implies that the center of the circumcircle of $\triangle ABC$ is midpoint of $BC$.
   Let $E$ be the second intersection of line $AD$ with the circumcircle of $\triangle ABC$.
   $\angle BAX=\angle BCX$, $\angle XAE=\angle XCE$, $\angle BAE=\angle BCE$ because subtend arcs $BX,\,XE,\,BE$ respectively. $\angle BEC=90^\circ$ as subtending half of the circle, $=\angle BDA$. Thus $\angle ABD=\angle CBE$.
   Let $M'=BE\cap CX$, then $\triangle BAM\sim \triangle BCM'$, $\triangle MAD\sim \triangle M'CE$, $\triangle BAD\sim \triangle BCE$, thus $\frac{BM}{MD}=\frac{BM'}{M'E}$ i.e. $$BM'=M'E.$$ More, as $\triangle BCA\sim \triangle BCE$ and it's quite evident that $AD=DE$ because chord $AE$ is perpendicular to diameter $BC$, thus $\triangle BCA$ and $\triangle BCE$ are congruent.

2. Note that $K$ lies on the perpendicular bisector of $BX$ (let $P$ be intersection of $OK$ with $BX$) because $\triangle PBO\sim\triangle XBC$ ($OK||XC$ and $\angle B$ is shared) thus $\frac{BX}{BP}=\frac{BC}{BO}=2$ hence $BP=PX$. $\angle BPO=\angle BXC=90^\circ$ because the latter sbtends half of the circle.
   Pass the line, parallel to $BX$ through $E$ and let $N,\,K'$ be the points of intersection of the line with $CX$ and $CB$ respectively. (Here is missing not the obvious step of proving $XK'=2XK$)
   $\triangle BXM'\sim\triangle ENM'$ because $EK'||BX$ and $\angle M'$ are vertical, but $BM'=M'E$ therefore $\triangle BXM'$ and $\triangle ENM'$ are congruent.
   ... this step should end with $$EN=NK'$$

3. $T$ is on perpendicular bisector to $XK'$ implies $TX=TK'$. $T$ is on perpendicular bisector to $EK'$ implies $TE=TK'$ thus $TX=TE$ and $T$ is on perpendicular bisector of $XE$.

4. Angle chasing. $\angle OKT=\angle KTX$ as $XC||KO$. In $\triangle KTX$ $\quad\angle K=90^\circ$ therefore $\angle KTX=90^\circ-\angle KXT$. $\angle KXT=\angle AXC$ subtends the same arc $AC$ as $\angle ABC$ therefore $\angle OKT=90^\circ-\angle ABC=\angle BCA$.
   Consider quadrilaterals $OPXQ$ (where $Q$ is midpoint of $XE$) and $BXEC$ -- both are inscribed and share angle $X$ thus $\angle KOT=\angle POQ=\angle BCE$, but $\angle BCE=\angle BCA$ thus $\angle KOT=\angle OKT$ thus $TO=TK$, QED.

Answer 3

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.