43. To inscribe a square in a given triangle.
Let ABC be the given triangle, and suppose DEFG the required square. Draw AH perpendicular to EC, and AK parallel to EG; and let EF produced meet AK at K Then BG is to GF as BA is to AK, and EG is to GD as BA is to AH{VI. 4). But GF is equal to GD, by hypothesis. Therefore BA is to AK as BA is to AH (V. 7, V. 11). Therefore AH is equal to AK{V. 7).
Hence we have the following synthetical solution. Draw AK parallel to EG, and equal to AH; and join BK. Then BK meets AC at one of the corners of the required square, and the solution can be completed.
44. Through a given point between two given straight lines, it is required to draw a straight line, such that the rectangle contained by the parts between the given point and the given straight lines may be equal to a given rectangle.
Let P be the given point, and AE and AC the given straight lines; suppose MPN the required straight line, so that the rectangle MP, PN is equal to a given rectangle.
Produce AP to Q, so that the rectangle AP, PQ may be equal to the given rectangle. Then the rectangle MP, PN is equal to the
rectangle AP, PQ. Therefore a circle will go round AMQN {Note on III. .35). Therefore the angle PNQ is equal to the angle PAM {III. 21).
Hence we have the following synthetical solution. Produce AP to Q, so that the rectangle AP, PQ may be equal to the given rectangle; describe on PQ a segment of a circle containing an angle equal to the angle PAM; join P with a point of intersection of this circle and AC; the straight line thus drawn solves the problem.
