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Higher Geometry Homework Answers – pg. 181-182 3. For an isosceles triangle that is not equilateral, how many symmedians coincide with their isogonal conjugates? In order for this to happen, a median (isogonal conjugate of a symmedian) would have to coincide with the angle bisector drawn from the same vertex. This can only happen at the vertex which is not on the base of the isosceles triangle. Answer is 1. 4. Prove that the median to a side of a triangle bisects any segment in the triangle parallel to that side. Let AE be a median of ▵ ABC and FG is parallel to BC. ∢ ADF = ∢ AEB (corresponding angles of parallel lines). So, ▵ ADF ~ ▵AEB by (AA). Thus, FD/BE = AD/AE. The same argument shows that ▵ ADG ~ ▵ AEC, and we get AD/AE = DG/EC. So, FD/BE = DG/EC, but since BE = EC (median) we get FD = DG. 6. Could the symmedian point be outside the triangle? No. Since the medians must be inside the triangle, the symmedians must also be inside, so their intersection is inside as well. 10. Prove that the altitudes of a triangle are the isogonal conjugates of the circumradii to the vertices of the triangle. See Quiz # 8. 12. Prove that a segment joining the feet of the perpendiculars from a point on a symmedian to the adjacent sides of the triangle is perpendicular to the corresponding median. Solution: In the diagram below, ▵ ABE is given with median AD and symmedian AF. H and G are the feet of the perpendiculars drawn to adjacent sides AB and AE from an arbitrary point F on the symmedian. We wish to prove that line HG is perpendicular to median AD. Since the angles at H and G are right angles, the quadrangle HFGA is concyclic, so we may draw the circle containing these four points. Triangles AHF and AGI are similar right triangles since the median and symmedian are isogonal conjugates. Thus ∢ HFA = ∢ AIG. In the circle, ∢ HAF and ∢ HGF are both measured by ½ of arc HF, and so are equal to each other. So, ▵ AHF and ▵ GIJ are similar which makes ∢ GJI a right angle, giving the result.