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2. Again, since

{PCED)=(PDEC},

(309°, Cor.) .. PDEC and PBFA are harmonic ranges having P a corresponding point in each. Therefore DB, EF, and CA are concurrent, and AC and DB meet on the polar of P.

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:. If from any point two secants be drawn to a circle, the connectors of their points of intersection with the circle meet upon the polar of the first point.

3. Since O is on the polar of P, P is on the polar of O. But since Q is a point from which secants are drawn satisfying the conditions of 2, Q is on the polar of O.

.. PQ is the polar of O.

Now ABCD is a concyclic quadrangle whereof AC and BD are internal diagonals and PQ the external diagonal.

.. In any concyclic quadrangle the external diagonal is the polar of the point of intersection of the internal diagonals, with respect to the circumcircle.

4. Since Q is on the polar of P and also on that of O, therefore PO is the polar of Q, and POQ is a triangle selfconjugate with respect to the circle.

5. Let tangents at the points A, B, C, D form the circumscribed quadrangle USVT.

Then S is the pole of AB, and T of DC.

.. ST is the polar of P, and S and T are points on the line QO.

Similarly U and V are points on the line PO.

But XY is the external diagonal of USVT, and its pole is O, the point of intersection of DB and AC.

.. X and Y are points on the line PQ.

Hence, If tangents be drawn at the vertices of a concyclic quadrangle so as to form a circumscribed quadrangle, the internal diagonals of the two quadrangles are concurrent, and their external diagonals are segments of a common line; and the point of concurrence and the line are pole and polar with respect to the circle.

EXERCISES.

1. UOVP and SOTQ are harmonic ranges.

2. If DB meets the line PQ in R, XOR is a self-conjugate triangle with respect to the circle.

3. To find a circle which shall cut the sides of a given triangle harmonically.

4. QXPY is a harmonic range.

313°. Let S be a circle and A'P'B'Q' a harmonic range.

Taking any point O on the

circle and through it projecting rectilinearly the points A'P'R'Q' we obtain the system APBQ, which is called a harmonic system of points on the circle.

Now, taking O' any other point on the circle, O'. APBQ is also harmonic. For

LAOP=LAOP,

LPOB PO'B, etc.

.. Def.-Four points on a circle

P

form a harmonic system when their

Р

B

joins with any fifth point on the circle form a harmonic pencil.

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.. When four points form a harmonic system on a circle, the rectangles on the opposite sides of the normal quadrangle which they determine are equal.

Cor. 2. If O comes to A, the ray OA becomes a tangent at A.

:. When four points form a harmonic system on a circle, the tangent at any one of them and the chords from the point of contact to the others form a harmonic pencil.

314°. Let the axis of the harmonic range APBQ be a tan

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Then, considering Aa, Pp, Bb, etc., as fixed tangents, and A'P'B'Q' as any other tangent.

LAOP LA'OP', POB=LP'OB', etc., (116°, Ex. 1) the pencils O.APBQ and O.A'P'B'Q' are both harmonic, and A'P'B'Q' is a harmonic range.

:. When four tangents form a harmonic system to a circle, they intersect any other tangent in points which form a harmonic range.

Cor. 1. If the variable tangent coincides with one of the fixed tangents, the point of contact of the latter becomes one of the points of the range.

.. When four tangents form a harmonic system to a circle, each tangent is divided harmonically by its point of contact and its intersections with the other tangents.

EXERCISES.

1. Tangents are drawn at A, A' the end-points of a diameter, and two points P, B are taken on the tangent through A such that AB=2AP. Through P and B tangents are drawn cutting the tangent at A' in P' and B'. Then 2A'B'=A'P', and AA', PB', and BP' are concurrent. 2. Four points form a harmonic system on a circle. Then the tangents at one pair of conjugates meet upon the secant through the other pair.

3. If four tangents form a harmonic system to a circle, the point of intersection of a pair of conjugate tangents lies on the chord of contact of the remaining pair.

4. If four points form a harmonic range, their polars with respect to any circle form a harmonic pencil; and conversely.

SECTION III.

OF ANHARMONIC PROPERTIES.

315°. Let A, E, C and D, B, F be two sets of three col

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P, Q, O. To show that these points are collinear.

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.. the pencils O. ECQF and O. ECPF are equianharmonic, and having three rays in common the fourth rays must be in common, i.e., they can differ only by a straight angle, and therefore O, P, Q are collinear.

(Being the first application of anharmonic ratios the work is very much expanded.)

.. If six lines taken in order intersect alternately in two sets of three collinear points, they intersect in a third set of three collinear points.

Cor. 1. ABC and DEF are two triangles, whereof each has one vertex lying upon a side of the other.

If AB and DE are taken as corresponding sides, A and F are non-corresponding vertices. But, if AB and EF are taken as corresponding sides, A and D are non-corresponding vertices.

Hence the intersections of AB and EF, of ED and CB, and of AD and CF are collinear.

.. If two triangles have each a vertex lying upon a side of the other, the remaining sides and the joins of the remaining non-corresponding vertices intersect collinearly.

Cor. 2. Joining AD, BE, CF, ADBE, EBFC, and ADFC are quadrangles, and P, Q, O are respectively the points of intersection of their internal diagonals.

.. if a quadrangle be divided into two quadrangles, the points of intersection of the internal diagonals of the three quadrangles are collinear.

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