Def.—The circle through these nine points is called the “nine points circle” of the triangle. 5. The circumcircle of a triangle is the “nine points circle” of each of the four triangles formed by joining the centres of the inscribed and escribed circles. 6. The distances between the vertices of a triangle and its orthocentre are respectively the doubles of the perpendiculars from the circumcentre on the sides. 7. The radius of the “nine points circle” of a triangle is equal to half its circumradius. PROP. VI.—Problem. In a given circle (ABCD) to inscribe a square. Sol.—Draw any two diameters AC, BD at right angles to each other. Join AB, BC, CD, DA. ABCD is a square. Dem.—Let O be the centre. Then the four angles at O, being right angles, are equal. Hence the arcs on which they stand are equal [III. xxvi.], and hence the four chords are equal [III. xxix.]. Therefore the figure ABCD is equilateral. Again, because AC is a diameter, the angle ABC is right [III. xxxi.]. In like manner the remaining angles are right. Hence ABCD is a square. PROP. VII.—Problem. About a given circle (ABCD) to describe a square. Sol.—Through the centre O draw any two diameters at right angles to each other, and draw at the points A, B, C, D the lines HE, EF, FG, GH touching the circle. EFGH is a square. Dem.—Because AE touches the circle at A, the angle EAO is right [III. xviii.], and therefore equal to BOC, which is right (const.). Hence AE is parallel to OB. In like manner EB is parallel to AO; and since AO is equal to OB, the figure AOBE is a lozenge, and the angle AOB is right; hence AOBE is a square. In like manner each of the figures BC, CD, DA is a square. Hence the whole figure is a square. Cor.—The circumscribed square is double of the inscribed square. PROP. VIII.—Problem. In a given square (ABCD) to inscribe a circle. Sol.—Bisect (see last diagram) two adjacent sides EH, EF in the points A, B, and through A, B draw the lines AC, BD, respectively parallel to EF, EH; then O, the point of intersection of these parallels, is the centre of the required circle. Dem.—Because AOBE is a parallelogram, its opposite sides are equal; therefore AO is equal to EB; but EB is half the side of the given square; therefore AO is equal to half the side of the given square; and so in like manner is each of the lines OB, OC, OD; therefore the four lines OA, OB, OC, OD are all equal; and since they are perpendicular to the sides of the given square, the circle described with O as centre, and OA as radius, will be inscribed in the square. PROP. IX.—Problem. About a given square (ABCD) to describe a circle. Sol.—Draw the diagonals AC, BD intersecting in O (see diagram to Proposition vi.). O is the centre of the required circle.