2001 JBMO Problems/Problem 2
Revision as of 23:27, 13 August 2018 by Rockmanex3 (talk | contribs) (Solution to Problem 2 (part 2 credit to efang) -- one easier and one harder)
Problem
Let be a triangle with and . Let be an altitude and be an interior angle bisector. Show that for on the line we have . Also show that for on the line we have .
Solution
Assume that . Since is an angle bisector of Thus, by AAS Congruency, which results in But so by proof by contradiction,
Assume that . Draw points and on and respectively such that is a rectangle. That means and are cyclic quadrilaterals, which means that
Because is bisected by , by the Angle Bisector Theorem, we have so by SAS Congruency, we have making and a square. This also means that so by AAS Congruency, making However, it is given that so by proof by contradiction.
See Also
2001 JBMO (Problems • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 | ||
All JBMO Problems and Solutions |