Auxillary circle and power of point

In $\triangle ABC$, the points $D$ and $E$ are located on $\overline{BC}$ such that $\angle BAD \cong \angle CAE$. Prove that $$\frac{AB^2}{AC^2}=\frac{BD\cdot BE}{CD\cdot CE}$$

Since the condition to prove looks a bit like the power of a point theorem, I tried constructing the circumcircle of $\triangle ADE$. The condition is easy to prove assuming that $\overline{AB}$ and $\overline{AC}$ are tangents to the circumcircle, but how do I prove that they are tangents? Since I haven't yet used the fact that $\angle BAD$ and $\angle CAE$ are equal, can it be proven using that condition?

• Took me a while to figure out that there was no point $Q$ in the triangle. – Arnaud Mortier Jun 14 '18 at 0:59

3 Answers

This is a straight forward application of the generalized angle bisector theorem. I am not sure how you can have AB and AC tangential to the circum-circle of triangle ADE unless you have angle BAC = 180 degrees which means ABC is not a triangle. Here is a sketch of the proof of your original proposition.

1. Apply the generalized angle bisector theorem to triangle ABC with point D on BC ignoring point E. Then you get: $\frac{BD}{DC} = \frac{AB\ sin(BAD)}{AC\ sin(DAC)}$

2. Apply the theorem again to triangle ABC with point E on BC ignoring point D. Then you get: $\frac{CE}{BE} = \frac{AC\ sin(CAE)}{AB\ sin(BAE)}$

3. Since angle BAD = angle CAE and angle BAE = angle DAC, we can equate 1 & 2 and rearrange terms to yield the required result.

Via the area method: we have $$\frac{BD}{CE} = \frac{[BAD]}{[CAE]} = \frac{AB \cdot AD}{AC \cdot AE}$$ where $[XYZ]$ denotes the area of $\triangle XYZ$. The first equation holds since $\triangle BAD$ and $\triangle CAE$ have the same height from vertex $A$, the second holds since they have the same angle at vertex $A$. More explicitly, if $H$ is the foot of the perpendicular from $A$, $$\frac{BD}{CE} = \frac{\frac12 \cdot BD \cdot AH}{\frac12 \cdot CE \cdot AH} = \frac{[BAD]}{[CAE]} = \frac{\frac12 \cdot AB \cdot AD \cdot \sin \angle BAD}{\frac12 \cdot AC \cdot AE \cdot \sin \angle CAE} = \frac{AB \cdot AD}{AC \cdot AE}.$$

For the exact same reason, we have $$\frac{BE}{CD} = \frac{[BAE]}{[CAD]} = \frac{AB \cdot AE}{AC \cdot AD}.$$ Multiply these together, and you get $$\frac{BD}{CE} \cdot \frac{BE}{CD} = \frac{AB^2}{AC^2}.$$

The condition is easy to prove assuming that $\overline{AB}$ and $\overline{AC}$ are tangents to the circumcircle

Point $A$ lies on the circle, so it is not possible that both $AB$ and $AC$ be tangents.

Alt. hint:   let $\angle BAD = \theta$ then applying the law of sines a few times:

\begin{align} \frac{BD}{\sin (\theta)} &= \frac{AB}{\sin(C+A-\theta)} \tag{1}\\[5px] \frac{BE}{\sin (A-\theta)} &= \frac{AB}{\sin(C+\theta)} \tag{2} \\[5px] \frac{CE}{\sin (\theta)} &= \frac{AC}{\sin(B+A-\theta)} \tag{3} \\[5px] \frac{CD}{\sin (A-\theta)} &= \frac{AC}{\sin(B+\theta)} \tag{4} \\[5px] \end{align}

It then follows by calculating $\,\frac{(1)\,\cdot\,(2)}{(3)\,\cdot\,(4)}\,$ that:

$$\require{cancel} \frac{BD\cdot BE}{CD\cdot CE} = \frac{AB^2}{AC^2}\cdot \frac{\bcancel{\sin(B+A-\theta)} \cdot \cancel{\sin(B+\theta)}}{\cancel{\sin(C+A-\theta)} \cdot \bcancel{\sin(C+\theta)}}$$