Question

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.

Answer 1

Answer:

Step-by-step explanation:

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.