In the convex quadrilateral $ABCD$, point $X$ is selected on side $AD$, and the diagonals intersect at point $E$. It is known that $AC = BD$, $\angle ABX = \angle AX B = 50^o$, $\angle CAD = 51^o$, $\angle AED = 80^o$. Find the value of angle $\angle AXC$.
Problem
Source: 2023 South Russian Girls MO - Assara Juniors p2
Tags: geometry, sngles
Davsch
28.02.2024 15:13
[asy][asy]
import olympiad;
pair A=(0,0);
pair C=dir(110);
pair D=dir(110+51)*1.1;
pair B=D+dir(110-80);
pair E=intersectionpoints(A--C,B--D)[0];
draw(A--B--C--D--A);
draw(A--C,green);
draw(B--D,green);
draw(A--B,blue);
dot(A);dot(B);dot(C);dot(D);dot(E);
real l=1-1/1.1;
pair X=A*l+(1-l)*D;
dot(X);
draw(A--X,blue);
draw(B--X);
label("$A$",A,SE);
label("$B$",B,SE);
label("$C$",C,NE);
label("$D$",D,SW);
label("$E$",E,S);
label("$X$",X,S);
markscalefactor=0.01;
draw(anglemark(D,B,A));
draw(anglemark(C,A,X));
[/asy][/asy]
Note that $\angle BAX=180^\circ-\angle XBA-\angle AXB=80^\circ$ and $AB=AX$ since $\angle XBA=50^\circ=\angle AXB$. Similarly, we have $\angle ADE=180^\circ-\angle EAD-\angle DEA=180^\circ-51^\circ-80^\circ=49^\circ$. Finally, $\angle BXD=\angle XBA+\angle BAX=130^\circ$, so $\angle DBX=180^\circ-\angle XDB-\angle BXD=1^\circ$.
This implies $\angle DBA=\angle DBX+\angle XBA=51^\circ=\angle CAX$. Hence, the triangles $ABD$ and $ACX$ are congruent, implying $\angle AXC=\angle BAD=80^\circ$.