MathDB

In a quadrilateral

ABCD

the intersection of the diagonals is called

P

. Point

X

is the orthocentre of triangle

PAB

. (The orthocentre of a triangle is the point where the three altitudes of the triangle intersect.) Point

Y

is the orthocentre of triangle

PCD

. Suppose that

X

lies inside triangle

PAB

and

Y

lies inside triangle

PCD

. Moreover, suppose that

P

is the midpoint of line segment

XY

. Prove that

ABCD

is a parallelogram. [asy] import geometry;
unitsize (1.5 cm);
pair A, B, C, D, P, X, Y;
A = (0,0); B = (2,-0.5); C = (3.5,2.2); D = A + C - B; P = (A + C)/2; X = orthocentercenter(A,B,P); Y = orthocentercenter(C,D,P);
draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw(A--extension(A,X,B,P), dotted); draw(B--extension(B,X,A,P), dotted); draw(P--extension(P,X,A,B), dotted); draw(C--extension(C,Y,D,P), dotted); draw(D--extension(D,Y,C,P), dotted); draw(P--extension(P,Y,C,D), dotted);
dot("

A

", A, W); dot("

B

", B, S); dot("

C

", C, E); dot("

D

", D, N); dot("

P

", P, E); dot("

X

", X, NW); dot("

Y

", Y, SE); [/asy]

Problem Statement