Let $ABC$ be a triangle, having no right angles, and let $D$ be a point on the side $BC$. Let $E$ and $F$ be the feet of the perpendiculars drawn from the point $D$ to the lines $AB$ and $AC$ respectively. Let $P$ be the point of intersection of the lines $BF$ and $CE$. Prove that the line $AP$ is the altitude of the triangle $ABC$ from the vertex $A$ if and only if the line $AD$ is the angle bisector of the angle $CAB$.
Problem
Source: Romanian Junior BkMO TST 2004, problem 5
Tags: trigonometry, geometry, angle bisector, geometry solved
Valentin Vornicu
14.05.2004 01:59
well, I just realised that this is the same problem as JBkMO 2000, problem 3!
Anonymous
22.05.2004 22:21
Maybe someone could post a solution also?
Peter Scholze
02.07.2004 15:10
hint for the guest: use ceva and do a short calculation. Peter
darij grinberg
07.07.2004 15:18
Actually, the solution with Ceva and trigonometry is clear. I'm just seeing there is a very simple solution without trigonometry: Let X be the foot of the A-altitude of triangle ABC. Then, we must prove that the lines AX, BF and CE concur if and only if the point D lies on the internal angle bisector of the angle CAB. Well, the triangles ABX and DBE are similar (since < ABX = < DBE and < AXB = < DEB = 90°), thus $\displaystyle \frac{BE}{DE}=\frac{BX}{AX}$, so that $\displaystyle BE=DE\cdot \frac{BX}{AX}$, and similarly $\displaystyle CF=DF\cdot \frac{CX}{AX}$. Therefore, $\frac{CF}{BE}=\frac{DF\cdot CX}{DE\cdot BX}$, and thus $\frac{AE}{BE}\cdot \frac{BX}{CX}\cdot \frac{CF}{AF}=\frac{AE}{AF}\cdot \frac{BX}{CX}\cdot \frac{CF}{BE}$ $=\frac{AE}{AF}\cdot \frac{BX}{CX}\cdot \frac{DF\cdot CX}{DE\cdot BX}=\frac{AE}{AF}\cdot \frac{DF}{DE}=\frac{AE: DE}{AF: DF}$. Now, by Ceva, the lines AX, BF and CE concur if and only if $\displaystyle \frac{AE}{BE}\cdot \frac{BX}{CX}\cdot \frac{CF}{AF}=1$. As we have seen, this is equivalent to $\displaystyle \frac{AE: DE}{AF: DF}=1$, or just $\displaystyle \frac{AE}{DE}=\frac{AF}{DF}$. But this holds if and only if the triangles ADE and ADF are similar (since we know < AED = < AFD = 90°), what, in turn, is equivalent to < EAD = < FAD, i. e. to the assertion that the point D lies on the angle bisector of the angle CAB. This completes the solution. EDIT: The same problem has been discussed on http://www.mathlinks.ro/Forum/viewtopic.php?t=41286 . Darij
Virgil Nicula
24.11.2005 02:53
Here is (the Darij's expression) "a clear, short and trigonometrical" solution (in my opinion): $x=m(\widehat {BAD})$, $y=m(\widehat {CAD})\Longrightarrow \frac{DB}{DC}=\frac{AB\sin x}{AC\sin y};$ $\ R\in AP\cap BC$ $\Longrightarrow\frac{RB}{RC}\cdot \frac{FC}{FA}\cdot \frac{EA}{EB}=1.$ $AP\perp BC\Longleftrightarrow \frac{RB}{RC}=\frac{AB\cos B}{AC\cos C}\Longleftrightarrow\frac{FA}{FC}\cdot \frac{EB}{EA}=\frac{AB\cos B}{AC\cos C}\Longleftrightarrow$ $(FA=AD\cos y,\ FC=DC\cos C,\ EB=DB\cos B,\ EA=AD\cos x)$ $\frac{DB}{DC}\cdot \frac{\cos y}{\cos x}=\frac{AB}{AC}\Longleftrightarrow \frac{AB}{AC}\cdot \frac{\sin x}{\sin y}\cdot \frac{\cos y}{\cos x}=\frac{AB}{AC}\Longleftrightarrow \tan x=\tan y\Longleftrightarrow$$x=y.$ Therefore, $AP\perp BC\Longleftrightarrow x=y\Longleftrightarrow$the ray $[AD$is the bisector of the angle $\widehat {BAC}$.