Parallelogram Isogonality Lemma
by liberator, Apr 12, 2016, 1:43 PM
Well I thought that I better make a blog post within the year so here it is. 
This nice lemma allows us to solve a variety of olympiad geometry problems. I first encountered it as BrMO2 2013/2. It is a generalisation of A Simple Lemma.
Lemma. The point
lies inside triangle 
so that 
. The point 
is such that 
is a parallelogram. Prove that 
.
![[asy]
unitsize(2.8cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8);
pointfontpen=fontsize(10pt);
real x=0.54;
pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P, Ap=A+B-P;
D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); DPA(B--P--C--Q--cycle^^Ap--A^^Ap--B^^Ap--Q);
/* Angle marks */
DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange);
/* Dots and labels */
D("A",A,dir(A));
D("B",B,dir(B));
D("C",C,dir(C));
D("P",P,dir(130));
D("Q",Q,dir(Q));
D("A'",Ap,dir(Ap));
[/asy]](//latex.artofproblemsolving.com/4/b/9/4b96e7c03be5d4af640ba4618a4e66fa034a772f.png)
Proof. Let
be such that 
are parallelograms. Then 
, so 
is cyclic. But then 
, as required. 
Some example problems:
Problem 1: Let
be a cyclic quadrilateral whose diagonals 
and 
meet at 
. The extensions of the sides 
and 
beyond 
and 
meet at 
. Let 
be the point such that 
is a parallelogram, and let 
be the image of under reflection in . Prove that 
are concyclic.
[ISL 2012 G2]
Problem 2: Let be a point inside triangle , and suppose lines 
, 
, 
meet the circumcircle again at 
, 
, 
(here 
, 
, 
). Let 
be any point in the interior of 
. A line through parallel to 
meets 
at 
, and the line through parallel to meets 
again at 
. Finally, the line through parallel to and the line through 
parallel to intersect at point . Given that 
and 
are parallel, prove that 
.
[Taiwan TST2 2014 Problem 6]
Problem 3: Let be a cyclic quadrilateral, and let diagonals and intersect at 
.Let 
and 
be the midpoints of segments 
and 
, respecctively. Lines 
and 
intersect at 
, and line 
intersects diagonals and at different points and , respectively. Prove that line 
is tangent to the circle through 
and .
[EGMO 2016 Problem 2]
This nice lemma allows us to solve a variety of olympiad geometry problems. I first encountered it as BrMO2 2013/2. It is a generalisation of A Simple Lemma.
Lemma. The point
lies inside triangle 
so that 
. The point 
is such that 
is a parallelogram. Prove that 
.![[asy]
unitsize(2.8cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8);
pointfontpen=fontsize(10pt);
real x=0.54;
pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), Q=B+C-P, Ap=A+B-P;
D(unitcircle,heavygreen); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--P,purple); DPA(B--P--C--Q--cycle^^Ap--A^^Ap--B^^Ap--Q);
/* Angle marks */
DPA(anglemark(P,B,A,5)^^anglemark(A,C,P,5),orange);
/* Dots and labels */
D("A",A,dir(A));
D("B",B,dir(B));
D("C",C,dir(C));
D("P",P,dir(130));
D("Q",Q,dir(Q));
D("A'",Ap,dir(Ap));
[/asy]](http://latex.artofproblemsolving.com/4/b/9/4b96e7c03be5d4af640ba4618a4e66fa034a772f.png)
Proof. Let
be such that 
are parallelograms. Then 
, so 
is cyclic. But then 
, as required. 
Some example problems:
Problem 1: Let
be a cyclic quadrilateral whose diagonals 
and 
meet at 
. The extensions of the sides 
and 
beyond 
and 
meet at 
. Let 
be the point such that 
is a parallelogram, and let 
be the image of under reflection in . Prove that 
are concyclic.[ISL 2012 G2]
![[asy]
unitsize(2.5cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8);
pointfontpen=fontsize(10pt);
pair A=dir(50), B=dir(105), C=dir(195), D=dir(-15), E=IP(D(A--C),D(B--D)), F=extension(A,D,B,C), G=C+D-E, H=reflect(A,D)*E;
DPA(A--B--C--D--cycle^^E--C--G--D--cycle^^D--H--F^^A--F--B);
D(E--F--G,purple);
D(unitcircle,heavygreen);
D(circumcircle(F,G,H),red+dashed);
/* Angle marks */
DPA(anglemark(A,C,F,5)^^anglemark(F,D,B,5),orange);
/* Dots and labels */
D("A",A,dir(A));
D("B",B,dir(B));
D("C",C,dir(C));
D("D",D,dir(D));
D("E",E);
D("F",F,dir(F));
D("G",G,dir(G));
D("H",H,dir(H));
[/asy]](http://latex.artofproblemsolving.com/9/f/9/9f918208e8639b1b8794bb25d4dd681dd540ee6c.png)
,
are isogonal. Thus 
so the result follows.Problem 2: Let
be a point inside triangle , and suppose lines 
, 
, 
meet the circumcircle again at 
, 
, 
(here 
, 
, 
). Let 
be any point in the interior of 
. A line through parallel to 
meets 
at 
, and the line through parallel to meets 
again at 
. Finally, the line through parallel to and the line through 
parallel to intersect at point . Given that 
and 
are parallel, prove that 
.[Taiwan TST2 2014 Problem 6]
![[asy]
unitsize(3.2cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8);
pointfontpen=fontsize(10pt);
/* Positions of pairs X,U */
real x=0.54, u=2/3;
pair sipc(pair A=(0,0), pair P, path c=unitcircle,real ext=10) {
return IP(L(A,P,-1/ext,ext),c);
}
pair A=dir(110), B=dir(205), C=dir(-25), X=WP(C--A,x), Y=IP(circumcircle(B,C,X),A--B), P=IP(B--X,C--Y), T=sipc(A,P), S=sipc(B,P), R=sipc(C,P), U=WP(P--T,u), V=extension(B,P,U,U+C-A), W=extension(C,P,U,U+B-A), Q=B+C-P;
filldraw(U--V--W--cycle,palegrey,pathpen);
filldraw(A--X--Y--cycle,palegrey,invisible);
D(unitcircle,heavygreen); D(circumcircle(B,C,X),red); D(A--B--C--cycle,heavygrey+linewidth(1)); D(Q--A--T,purple); DPA(B--Q--C^^X--Y^^R--S); DPA(B--S^^C--R,pathpen+linetype("4 4")+linewidth(1));
/* Angle marks */
DPA(anglemark(X,B,A,5)^^anglemark(X,C,Y,5),orange);
/* Dots and labels */
D("A",A,dir(A));
D("B",B,dir(B));
D("C",C,dir(C));
D("X",X,dir(75));
D("Y",Y,dir(100));
D("P",P,dir(120));
D("T",T,dir(T));
D("S",S,dir(S));
D("R",R,dir(R));
D("U",U,dir(U));
D("V",V,N);
D("W",W,SSE);
D("Q",Q,dir(Q));
[/asy]](http://latex.artofproblemsolving.com/4/c/5/4c5faf01235c9980f2376002ba9ca025eba12b57.png)
.
are in perspective, so it follows by Desargue's theorem that 
is at infinity: i.e. 
.
is cyclic by Reim's theorem.
,Problem 3: Let
be a cyclic quadrilateral, and let diagonals and intersect at 
.Let 
and 
be the midpoints of segments 
and 
, respecctively. Lines 
and 
intersect at 
, and line 
intersects diagonals and at different points and , respectively. Prove that line 
is tangent to the circle through 
and .[EGMO 2016 Problem 2]
![[asy]
unitsize(3cm);
pointpen=black; pathpen=rgb(0.4,0.6,0.8);
pointfontpen=fontsize(10);
pair A=dir(-140), B=dir(-40), C=dir(75), D=dir(170), X=IP(D(A--C),D(B--D)), C1=(X+C)/2, D1=(X+D)/2, Y=extension(A,D1,B,C1), M=(C+D)/2, E=extension(Y,M,A,C), F=extension(Y,M,B,D);
D(A--B--C--D--cycle,pathpen+linewidth(1));
DPA(C1--M--D1--X--cycle^^A--Y--B);
D(X--Y--F,purple);
D(unitcircle,heavygreen);
D(circumcircle(X,E,F),red);
D(circumcircle(A,B,C1),dashed+pink);
D("A",A,dir(A));
D("B",B,dir(B));
D("C",C,dir(C));
D("D",D,dir(D));
D("E",E,NE);
D("F",F,dir(F));
D("M",M,dir(150));
D("X",X,dir(X));
D("Y",Y,dir(Y));
D("C_1",C1,dir(0));
D("D_1",D1,SW);
[/asy]](http://latex.artofproblemsolving.com/c/c/f/ccf1f23917419124870a1a4899c62c35684c8e9d.png)
is cyclic. Note that 
,
are isogonal wrt 
.
,
,
.This post has been edited 1 time. Last edited by liberator, Apr 15, 2016, 8:21 PM
Reason: added EGMO 2016/2
Reason: added EGMO 2016/2
October 2018
April 2018