Library / Croatian Mathematical Competitions /84 of 114

Croatian Mathematical Competitions

Geometry Difficulty 6.2 National Olympiad Prove it Croatia

Let $ABC$ be a triangle such that $|AB| > |AC|$ . Let $P$ be the midpoint of the segment $\overline{BC}$ and $S$ the intersection of the bisector of the angle $\angle BAC$ and the segment $\overline{BC}$ . The line parallel to the line $AS$ , which passes through $P$ , intersects the lines $AB$ and $AC$ in points $X$ and $Y$ respectively. Let $Z$ be a point such that $Y$ is the midpoint of the segment $XZ$ , and that lines $BY$ and $CZ$ intersect at $D$ .
Prove that the bisector of the angle $\angle BDC$ is parallel to the line $AS$ .
(D. Monk, New Problems in Euclidean Geometry)

Solution

Since $|BP| = |CP|$ and
$\frac{|BX|}{|BP|} = (\text{similarity}) = \frac{|BA|}{|BS|} = (\text{angle bisector}) = \frac{|CA|}{|CS|} = (\text{similarity}) = \frac{|CY|}{|CP|},$
we conclude that $|BX| = |CY|$ .

Note that $\triangle XYA = \triangle SAC = \triangle BAS = \triangle BXP = \triangle AXY$ , and therefore $\triangle YXB = \triangle CYZ$ .
Since $|XY| = |YZ|$ and $|BX| = |CY|$ , the SAS congruence theorem implies that the triangles $BXY$ and $CYZ$ are congruent.
Hence $\triangle BYP = \triangle PZC$ , i.e. $\triangle DYZ = \triangle YZD$ . Since $\triangle DYZ + \triangle YZD = \triangle BDC$ , it follows that the bisector of the angle $\triangle BDC$ is parallel to the line $ZP$ , and hence to the line $AS$ as well.

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.