<s>[INST] Prove that $TK=TO$ 
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

 [/INST] 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.
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