SELECTION and TRAINING SESSION

Question

Let ABCD be a cyclic quadrilateral, P = AB CD, Q = AD BC. Prove that the distance between the orthocenters of the triangles APD and AQB is equal to that of the triangles CQD and BPC. FIGURE_0

Accepted Answer

We refer to the following well-known fact: in any cyclic quadrilateral all perpendiculars drawn through the midpoints of the sides to the opposite sides meet at the same point

U

. It follows that the altitudes

D D_{1}

and

C C_{1}

of the triangle

A D B

and

B C P

, respectively, are symmetric with respect to

U

. The same is true for the altitudes

A A_{1}

and

B B_{1}

. Therefore the intersection points

D D_{1} \cap A A_{1}

and

C C_{1} \cap B B_{1}

are symmetric with respect to

U

. That is, the orthocenters

H_{A D P}

and

H_{B C P}

are symmetric. Similarly, the orthocenters

H_{A QB}

and

H_{D QC}

are symmetric with respect to

U

. So the segments

H_{A D P} H_{A QB}

and

H_{C Q D} H_{B C P}

are symmetric, so they have the same length.