Solution of a Geometrical Problem

R. F. Muirhead
1903 Proceedings of the Edinburgh Mathematical Society  
To draw through a given point a transversal of a given triangle so that the segments of the transversal may be in a given ratio." FIGURE 25. Analysis. ABC is a triangle and DEF a transversal and K is the point of concurrence of the four circles circumscribed about the four triangles formed by the transversal and the sides of the triangle. H is a point on the circumference of the circumcircle of ABC, such that AK and AH are equally inclined to the bisector of the angle BAC. Thus z.HBC = . L K C
more » ... us z.HBC = . L K C B = â nd ^HOB= i K B C = ^K F E . Thus the triangles HBC, KCB and KEF are mutually similar. Again if AH meet BC in X, we havê .XHC= <_ABC= ^FKD. Hence the figures HBCX and KEFD are similar ; hence CX : XB = FD : DE. Let O be any point in EF, and OLDK a circle meeting KB in L. Then L OLK = L. ODK = L FBK. .-. OL is parallel to FB. Construction. Given ABC and 0, and the ratio p:q to which FD:DE is to be equal. Make GX:XB=p:q. Draw AXH meeting the circumcircle of ABC in H. Make arc BK equal and opposite to arc CH. Draw OL parallel to BA to meet KB in L. and let the circle through OLK cut BC in D and D'. Then the lines OD and OD' give the two solutions of the problem.
doi:10.1017/s0013091500001991 fatcat:d263e6x2xfdvjcux7l74uvipoi