My problem is:

$AB$ is a diameter of a circle and $BM$ is the tangent at $B$. If the tangent at $C$ on the circle meets $BM$ at $X$ and if $AC$ produced meets $BM$ at $Y$, prove $BX=XY$.

And this was how I approached, In the figure, we join $B$ to $C$ to have angle $ACB$ equal to $90^{\circ}$. and see that $XB=XC$. Also, that angle $ABY$ equals to $90^{\circ}$.
But, I am not able to have $XY$ anywhere. All, I see and have is a few angles and as sides only radii and the above relation between $XB$ and $XC$.

If anybody could tell me how can I have $XY$ or anything related to that in my solution or any other way to approach this problem, (I would prefer this rather than solution) I will really appreciate that.

up vote 2 down vote accepted

The diagram below shows my understanding of your description.

I have added Blue colored items (dashed lines and texts). Lengths can be deduced readily.

You can see that the angle OCA and the angle YCX are equal.

Also, CX = CT

As a result the two triangles CXY and COA are the same.

It follows that YX=AT, and hence x is the middle of BY.

enter image description here

Let $O$ be the center of the circle. Then $OBX$ and $OCX$ are congruent by symmetry and hence $\angle BOX = \frac12 \angle BOC$. Since also $\angle BOC = 2\angle BAC$, triangle $OBX$ is half the size of $ABY$.

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