How to rewrite this problem in AlphaGeometry language?

[mathinfinity]

This is not a hard problem. I have solved this but I am not able to translate this into AlphaGeometry language.
ABC is a triangle with incentre I. The feet of the altitudes from I to BC, AC, AB are D, E, F respectively, and the line through D parallel to AI intersects AB and AC at X and Y respectively. Prove that the circles with diameters XF and YE have a common point on the circumcircle of ABC.

I have tried programming this but I have encountered some error here. This is my code.
a b c = triangle a b c; o = circle o a b c; i = incenter i a b c; d = foot d i b c; e = foot e i a c; f = foot f i a b; x = on_pline x d a b, on_line x a b; y = on_pline y d a c, on_line y a c; m1 = midpoint m1 x f; m2 = midpoint m2 y e; q = on_circle q m1 x, on_circle q m2 y ? cyclic q a c b

But when I run it, there are no responses from alphageometry and ddar (other example problem works tho), so I think it's some problem about the translation into alphageometry language. Can anyone help me?

[mathinfinity]

The problem has been resolved. It turns out that it's a careless mistake by myself. Thanks.

[foldl]

That's great. Could you share your updated code about this problem? Frankly speaking, I am still learning its language.

[mathinfinity]

It turns out that x=on_pline x d a b is wrong, it should be x=on_pline x d a i. It same goes with y.

[foldl]

Got it. Now I know on_pline means "on parallel line".

[sobiny]

Actually, the unresponsiveness issue isn’t caused by ddar—it’s caused by the line:

  g, _ = gh.Graph.build_problem(p, DEFINITIONS)

Whenever duplicate points are defined—for example, if both C and D are defined as the midpoint of AB—the construction will hang. I haven’t had the time to pinpoint exactly where the problem originates. Similarly, if you use the nine-point circle, the orthocenter is computed during its definition, and if the orthocenter is defined again later, it will also cause a freeze.

So, everyone can add a breakpoint at this location to check whether it’s really hanging here.

实际上无响应的问题还不是出在ddar上
是出在
  g, _ = gh.Graph.build_problem(p, DEFINITIONS)
这行代码上。
只要定义重复的点。比如 C是 AB的中点 D是AB的中点。在构建的时候就会卡死。我还没有时间去跟问题出在哪。同理的，如果使用了9点圆，在定义的时候就会做垂心，如果后面又定义了垂心，也会卡死。
所以，大家可以在 这个位置上，做一个标记，看到底是不是卡在了这里。

[sobiny]

I discovered that the issue stemmed from a geometry problem in this year’s second round of the Chinese IMO Training Team selection (50 to 16):

At first, I defined the following:

a b c = triangle; x y z i = ninepoints x y z i a b c; p = on_circle p i x; q = on_tline q p a p, on_line q b c; x0 = on_tline x0 a a q, on_line x0 p q; h = orthocenter h a b c; d = midpoint d b c; o = circumcenter o a b c; k = on_tline b a c, on_tline c a b ? perp a k b c

This was impossible—it would never run through. I kept debugging and eventually found that the error was caused by the last line of code.

Then, by removing each definition one by one, I discovered that the conflict arose when using
x y z i = ninepoints x y z i a b c
to compute the center of the nine-point circle, which clashed with either
h = orthocenter h a b c
or
o = circumcenter o a b c.

Finally, I constructed the statements in another way using a similar method:

a b c = triangle a b c; o = circumcenter o a b c; d = on_tline d b a c, on_tline d c a b; m = midpoint m o d; e = midpoint e b c; p = on_circle p m e; q = on_tline q p a p, on_line q b c; x = on_tline x a a q, on_line x p q; z = midpoint z a q ? perp x d e z
I first computed the triangle’s circumcenter, then determined the orthocenter d, and then used the midpoint between the circumcenter and the orthocenter to define the center of the nine-point circle. For perp x d e z, I did not redefine the triangle’s orthocenter; instead, I directly used the previously computed orthocenter d that was used for the nine-point circle. With this, the code ran successfully. Using your code, I was able to prove the problem from the domestic selection contest in over 120 steps.

我发现这个问题，还是因为今年的中国 IMO集训队选拔第二轮（50进16）有一道几何题：

我最先的时候定义：
a b c = triangle; x y z i = ninepoints x y z i a b c; p = on_circle p i x; q = on_tline q p a p, on_line q b c; x0 = on_tline x0 a a q, on_line x0 p q; h = orthocenter h a b c; d = midpoint d b c; o = circumcenter o a b c; k = on_tline b a c, on_tline c a b ? perp a k b c

不可能的，是永远跑不通的。我就一直在定位，然后发现是卡在上面那一行代码了。
然后我又通过删除每个定义的方式，然后发现是 我通过 x y z i = ninepoints x y z i a b c 九点圆求圆心的时候和 h = orthocenter h a b c 或者 o = circumcenter o a b c这句冲突了。
最后通过另外的定义的方式，用同样的方法来构造好了语句：
a b c = triangle a b c; o = circumcenter o a b c; d = on_tline d b a c, on_tline d c a b; m = midpoint m o d; e = midpoint e b c; p = on_circle p m e; q = on_tline q p a p, on_line q b c; x = on_tline x a a q, on_line x p q; z = midpoint z a q ? perp x d e z

我先求三角形的外心，然后求三角形的垂心d，然后外心和垂心的中点来定义九点圆的圆心。perp x d e z 这里，就不重新设置三角形的垂心了。直接用之前求九点圆时候用到的垂心d，然后代码顺利跑通了。用您的代码通过120多步证明了这道国内选拔赛的题。