In a triangle ABC, AB = AC. D is the mid-point of BC, E is the foot of the perpendicular drawn from D to AC, and F is the mid-point of DE.
Prove that AF is perpendicular to BE.
Nice proof Heureka, my first attempt was by vectors but I couldn't find one, I intended to get back to it :) , but I switched my attention to a co-ordinate geometry approach.
I have two such solutions. The first had its origin at D with DA as the vertical axis. That worked but the algebra was messy in places. Much better it turns out is to have A as the origin with AC along the x-axis. (Diagram below).
Let B and C have co-ordinates (p, q) and (c, 0) respectively, then D will have co-ordinates
((p + c)/2, q/2).
From there, E will be ((p + c)/2, 0) and F ((p + c)/2, q/4).
The slope of AF will be \(\displaystyle \frac{q}{4}.\frac{2}{p+c} = \frac{q}{2(p+c)}\)
and the slope of BE\(\displaystyle\frac{q-0}{p-(p+c)/2}=\frac{2q}{p-c}\).
The product of those is
\(\displaystyle\frac{q}{2(p+c)}.\frac{2q}{(p-c)}=\frac{q^{2}}{p^{2}-c^{2}}\),
which, since
\(p^{2}+q^{2}=c^{2}\text{ so that } p^{2}-c^{2}=-q^{2}\)
is equal to -1.
Hence, AF and BE are at right angles to each other.
- Bertie
+26367 In a triangle ABC, AB = AC. D is the mid-point of BC, E is the foot of the perpendicular drawn from D to AC, and F is the mid-point of DE. Prove that AF is perpendicular to BE.
I. Definition:
\(\boxed{~ \begin{array}{lcl} \vec{a}= \vec{BC}\qquad |\vec{a}|= a \\ \vec{b}= \vec{AC}\qquad |\vec{b}|= b\qquad \vec{b}\cdot \vec{b} = b^2 \\ \quad \vec{a}\cdot \vec{b} = \frac{a}{2}\cdot a\\ \quad \vec{a}\cdot \vec{b} = \frac{a^2}{2}\\ \hline \vec{d}=\vec{AD}\\ \vec{d}=\vec{b}-\frac12 \vec{a} \\ \end{array} ~} \boxed{~ \begin{array}{lcl} \vec{f}=\vec{AE}\\ \vec{f}= \frac{ (\vec{b}\cdot\vec{d}) } {b^2} \vec{b}\\ \vec{f}= \frac{ (\vec{b}\cdot \left(\vec{b}-\frac12 \vec{a} \right) ) } {b^2} \vec{b}\\ \vec{f}= \frac{ (b^2 - \frac{ (\vec{a}\cdot \vec{b}) }{2} ) } {b^2} \vec{b}\\ \vec{f}= \frac{ (b^2 - \frac{ a^2 }{4} ) } {b^2} \vec{b}\\ \vec{f}= \left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \end{array} ~}\\\)
\(\boxed{~ \begin{array}{lcl} \vec{e}=\vec{ED}\\ \vec{e}=\vec{d}-\vec{f}\\ \vec{e}=\left( \vec{b}-\frac12 \vec{a} \right) -\vec{f}\\ \frac{\vec{e}}{2}=\left( \vec{b}-\frac12 \vec{a} \right)\frac12 -\frac12 \vec{f}\\ \vec{AF} = \frac{\vec{e}}{2}+\vec{f} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 -\frac12 \vec{f}+\vec{f}\\ \vec{AF} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 +\frac12 \vec{f}\\ \vec{AF} = \left( \vec{b}-\frac12 \vec{a} \right)\frac12 +\frac12 \left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{AF} = \vec{b} \left[ \frac12 +\frac12 \cdot \left( 1 - \frac{ a^2 }{4b^2} \right) \right] -\frac14 \cdot \vec{a} \\ \vec{AF} = \vec{b} \left( \frac12 +\frac12 - \frac{ a^2 }{8b^2} \right) -\frac14 \cdot \vec{a} \\ \vec{AF} = \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) -\frac14 \cdot \vec{a} \\ \end{array} ~} \boxed{~ \begin{array}{lcl} \vec{BE} = \vec{a}-(\vec{b}-\vec{f})\\ \vec{BE} = \vec{a}-\vec{b}+\vec{f}\\ \vec{BE} = \vec{a}-\vec{b}+\left(1 - \frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{BE} = \vec{a}-\vec{b}+ \vec{b} - \left(\frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \vec{BE} = \vec{a} - \left(\frac{ a^2 }{4b^2} \right) \cdot \vec{b}\\ \end{array} ~}\)
II. Prove that AF is perpendicular to BE.
\(\boxed{~ \begin{array}{lcl} \vec{AF} = -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \\ \vec{BE} = \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right)\\ \text{Perpendicular, if } \ (~\vec{AF}\cdot \vec{BE}~) = 0 \\ \hline \\ (~\vec{AF}\cdot \vec{BE}~) = \left ( -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \right) \cdot \left( \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right) \right) \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac14 \cdot \vec{a} \cdot \vec{a} + \frac14 \cdot \vec{a} \cdot \vec{b} \left(\frac{ a^2 }{4b^2} \right) + \vec{b} \cdot \vec{a} \left( 1 - \frac{ a^2 }{8b^2} \right) - \vec{b}\cdot \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ \qquad \vec{a} \cdot \vec{a} = a^2 \qquad \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} =\frac{a^2}{2} \qquad \vec{b}\cdot \vec{b}=b^2 \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac14 \cdot \frac{a^2}{2} \left(\frac{ a^2 }{4b^2} \right) + \frac{a^2}{2} \left( 1 - \frac{ a^2 }{8b^2} \right) - b^2 \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} - \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} -\frac{ a^2 }{4}+ \frac{ a^4 }{32b^2} \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4}-\frac{ a^2 }{4}+ \frac{a^2}{2} + \frac{a^4}{32b^2} + \frac{ a^4 }{32b^2} - \frac{ a^4 }{16b^2} \\ \color{red }(~\vec{AF}\cdot \vec{BE}~) \color{black }= -\frac{ a^2 }{2}+ \frac{a^2}{2} + \frac{a^4}{16b^2} - \frac{ a^4 }{16b^2} \color{red }= 0\\ \end{array} ~}\)
Nice proof Heureka, my first attempt was by vectors but I couldn't find one, I intended to get back to it :) , but I switched my attention to a co-ordinate geometry approach.
I have two such solutions. The first had its origin at D with DA as the vertical axis. That worked but the algebra was messy in places. Much better it turns out is to have A as the origin with AC along the x-axis. (Diagram below).
Let B and C have co-ordinates (p, q) and (c, 0) respectively, then D will have co-ordinates
((p + c)/2, q/2).
From there, E will be ((p + c)/2, 0) and F ((p + c)/2, q/4).
The slope of AF will be \(\displaystyle \frac{q}{4}.\frac{2}{p+c} = \frac{q}{2(p+c)}\)
and the slope of BE\(\displaystyle\frac{q-0}{p-(p+c)/2}=\frac{2q}{p-c}\).
The product of those is
\(\displaystyle\frac{q}{2(p+c)}.\frac{2q}{(p-c)}=\frac{q^{2}}{p^{2}-c^{2}}\),
which, since
\(p^{2}+q^{2}=c^{2}\text{ so that } p^{2}-c^{2}=-q^{2}\)
is equal to -1.
Hence, AF and BE are at right angles to each other.
- Bertie
