+0

# 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.

0
1519
4

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.

Guest Nov 23, 2015

#2
+25

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

Guest Nov 25, 2015
Sort:

#1
+18948
+25

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.

$$\boxed{~ \begin{array}{lcl} \vec{AF} = -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \\ \vec{BE} = \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right)\\ \text{Perpendicular, if } \ (~\vec{AF}\cdot \vec{BE}~) = 0 \\ \hline \\ (~\vec{AF}\cdot \vec{BE}~) = \left ( -\frac14 \cdot \vec{a} + \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \right) \cdot \left( \vec{a} - \vec{b} \left(\frac{ a^2 }{4b^2} \right) \right) \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac14 \cdot \vec{a} \cdot \vec{a} + \frac14 \cdot \vec{a} \cdot \vec{b} \left(\frac{ a^2 }{4b^2} \right) + \vec{b} \cdot \vec{a} \left( 1 - \frac{ a^2 }{8b^2} \right) - \vec{b}\cdot \vec{b} \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ \qquad \vec{a} \cdot \vec{a} = a^2 \qquad \vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a} =\frac{a^2}{2} \qquad \vec{b}\cdot \vec{b}=b^2 \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac14 \cdot \frac{a^2}{2} \left(\frac{ a^2 }{4b^2} \right) + \frac{a^2}{2} \left( 1 - \frac{ a^2 }{8b^2} \right) - b^2 \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4b^2} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} - \left( 1 - \frac{ a^2 }{8b^2} \right) \left(\frac{ a^2 }{4} \right)\\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4} + \frac{a^4}{32b^2} + \frac{a^2}{2} - \frac{ a^4 }{16b^2} -\frac{ a^2 }{4}+ \frac{ a^4 }{32b^2} \\ (~\vec{AF}\cdot \vec{BE}~) = -\frac{ a^2 }{4}-\frac{ a^2 }{4}+ \frac{a^2}{2} + \frac{a^4}{32b^2} + \frac{ a^4 }{32b^2} - \frac{ a^4 }{16b^2} \\ \color{red }(~\vec{AF}\cdot \vec{BE}~) \color{black }= -\frac{ a^2 }{2}+ \frac{a^2}{2} + \frac{a^4}{16b^2} - \frac{ a^4 }{16b^2} \color{red }= 0\\ \end{array} ~}$$

heureka  Nov 25, 2015
#2
+25

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

Guest Nov 25, 2015
#3
+18948
+10

Very very Nice proof Berti !

heureka  Nov 26, 2015
#4
+82757
+5

Very nice, heureka and Bertie......!!!!!

CPhill  Nov 26, 2015

### 26 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details