For draw AC, and AP, and PC; and let AC

cut BD in E;

... (E. 38. 1.)

(S. 42. 1.) AE= EC.

▲ ABE = ACBE; and ▲ APE = A CPE; .. A APB A CPB;

(E. 41. 1.) ABX PKBC X PL; .. (E. 16. 6.) AB: BC:: PL: PK. Again, since (constr. E. 15. 1. E. 32. 1.) the ▲ BKP, BNQ, are equiangular, as are, also, the BLP, BMQ,

.. (E. 4. 6.) QM: BQ:: PL: PB;

and BQ: QN:: PB: PK,

.. (E. 22. 5.) QM: QN:: PL: PK:

And it has been proved that AB : BC:: PL: PK; .. (E. 11. 5.) AB: BC;: QN: QM.

In the same manner, also, the proposition may be shewn to be true, if perpendiculars be let fall from Q on the sides DA, and DC, produced.

PROP. LIV.

63. PROBLEM. From a given point, in the base of a scalene triangle, to draw a straight line, which shall cut off equal segments from the two remaining sides, the less of those sides having been produced.

Let D be a given point in the base BC of the

scalene ▲ ABC: It is required to draw from Da straight line which shall cut off from the greater side AC, and from the less side AB, produced, equal segments.

From AC cut off (E. 3. 1.) CE=AB; through D draw (E. 31. 1.) DF parallel to AB, and DG parallel to AC; and, accordingly as the point F falls between E and C, or between E and A, take in AG, or in GA produced, AH EF; produce HG, or GH, (S. 73. 3. cor.) to K, so that GK X KH=GA X AF; lastly, join D, K: Then shall DK cut off from AC a segment CL equal to the

segment BK, which it cuts off from AB produced. For, since (constr.) GK × KH is equal to GA × AF, or (constr. and E. 34. 1.) to DFX GD,

1

(E. 16. 6.) GK: GD:: DF: KH;

But (constr. E. 29. 1. and S. 26. 1.) the two A KGD, LFD, are equiangular;

.. (E. 4. 6.) GK:GD:: DF: FL:

.*. (E. 9. 5.) KH=FL:

But (constr.) BH=CF; to these equals add the equals HK, and FL, and it is manifest, that the segment CL is equal to the segment BK.

PROP. LV.

64. THEOREM. If an angle of a triangle be bisected by a straight line, which also cuts the base, the rectangle, contained by the sides of the triangle, is equal to the rectangle contained by the segments of the base, together with the square of the straight line bisecting the angle.

Let ABC be a A, and let the BAC be bisected by AD; then BAX AC=BD× DC+ AD*.

Describe (E. 5. 4.) the circle ACB about the triangle; produce AD to the circumference in E, and draw EC. And, because (E. 21. 3.) the ABC AEC, and (hyp.) the BAD = 4 CAE, ... (E. 32. 1.) the ▲ ABD, AEC, are equiangular;

.. (E. 4. 6.) BA: AD:: EA: AC;

.. (E. 16. 6.) BA × AC=EA× AD; i. e. (E. 3. 2.) BA × AC = ED × DA + AD': But (E. 35. 3.) ED X DA=BD X DC; .. BAX AC = BD X DC + AD2.

PROP. LVI.

65. THEOREM. If from any angle of a triangle a straight line be drawn perpendicular to the base, the rectangle contained by the sides of the triangle, is equal to the rectangle contained by the perpendicular and the diameter of the circle described about the triangle.

Let ABC be a A, and AD the from the BAC to the base BC; then is BA X AC equal to the rectangle contained by AD, and the diameter of the circle described about the ▲ ABC.

Describe (E. 5. 4.) the circle ACB about the

B

triangle; draw its diameter AE, and join E, C:
Because the ECA in a semi-circle is equal (E.
31. 3.) to the right ▲ BDA, and that (E. 21. 3.)
the AEC ABC, ... (E. 32. 1.) the
AEC are equiangular,

.. (E. 4. 6.) BA: AD:: EA: AC;
.. (E. 16. 6.) BAX AC=EA × AD.

ABD,

PROP. LVII.

66. THEOREM. The rectangle contained by the diagonals of a quadrilateral rectilineal figure, inscribed in a circle, is equal to both the rectangles contained by its opposite sides.

Let ABCD be any quadrilateral rectilineal figure, inscribed in a circle ACB, and let AC, BD, be its diagonals; then AC × BD = AB X CD + ADX BC.

Make (E. 23. 1.) the ABE DBC; add to each the common ▲ EBD; .. the ▲ ABD =