如何将此 Asymptote 代码渲染成图像？

Question

I don't know anything about asymptote, I asked an AI to generate the code based this geometry problem,

Prove that in a triangle ABC, if the altitudes AD, BE and CF to sides AB, AC and BC respectively, are concurrent, then the triangle is an isosceles triangle.

Here is the Asymptote code it generated,

pointpen = black;
pathpen = black+linewidth(0.7);
size(180);
pair A=(0,0),B=(1,0),C=IP(CR(A,3),CR(B,2)),D=foot(A,B,C),E=foot(B,A,C),F=foot(C,A,B);
D(MP("A",A,SW)--MP("B",B,SE)--MP("C",C,N)--cycle);
D(A--MP("D",D,S)--B);
D(B--MP("E",E,S)--C);
D(C--MP("F",F,S)--A);
D(D);D(E);
D(F);
D(circumcircle(A,B,C));
D(incircle(A,B,C));

I found this renderer online, http://asymptote.ualberta.ca/ . When I enter this into the renderer I get this error,

pointpen = black;
^
workspace_1.asy: 1.1: no matching variable 'pointpen'

Update: OK on further investigation seems like the AI may be using some sort of short-hand mapping or making assumptions about pre-existing variables. I think variables like pointpen don't really exist, and maybe this D stands for draw() . Will keep trying.

I am looking at Asymptote's tutorial here ( https://asymptote.sourceforge.io/asymptote_tutorial.pdf ), and I can't find any of these things like pointpen , pathpen , or these D things here, so this isn't standard Asymptote code? Searched high & low can't really figure out what this is or how to visualize it. Any help would be appreciated thanks!

Answer 1

OK finally figured this out. I think the AI was using some older (or alternative version) of Asymptote.

I think the modern equivalent of the code it generated looks like this,

size(180); 
pair A=(0,0),B=(1,0),C=(0.5,sqrt(3)/2),D=(A+B)/2,E=(B+C)/2,F=(A+C)/2;
draw(A--B--C--cycle);
draw(A--D); draw(B--E); draw(C--F);
dot("$A$",A,SW); dot("$B$",B,SE); dot("$C$",C,N); dot("$D$",D,S); 
dot("$E$",E,S); dot("$F$",F,S);

Of course I can't be 100% sure this is what the AI intended (does it have intentions?) but this graphic looks like it fits what the geometry problem is asking.

Sorry my original question was super vague.