Triangle Equality Proof: $\frac{a}{b}=1+\sqrt{2\left(\frac{c}{b}\right)^2-1}$

[anemone]

Let $ABC$ be a triangle with $\angle ACB=2\alpha,\, \angle ABC=3\alpha$, $AD$ is an altitude and $AE$ is a median such that $\angle DAE=\alpha$. If $BC=a,\,CA=b$ and $AB=c$, prove that $\dfrac{a}{b}=1+\sqrt{2\left(\dfrac{c}{b}\right)^2-1}$.

 

[jvicens]

Thank you for bringing up this interesting problem. I am always excited to explore new mathematical concepts and solve challenging problems.

First, let's draw a diagram of the given triangle $ABC$ with the labeled points $D$ and $E$ as shown below.

[insert diagram here]

We are given that $\angle ACB=2\alpha$ and $\angle ABC=3\alpha$. This means that the remaining angle $\angle BAC$ must be $180^\circ - 2\alpha - 3\alpha = 180^\circ - 5\alpha$.

Next, we know that $AD$ is an altitude, which means it is perpendicular to the base $BC$. This forms a right triangle $ADC$ with $\angle ACD=90^\circ$. Since $\angle DAE=\alpha$, we can use the fact that the sum of angles in a triangle is $180^\circ$ to find that $\angle EAD=90^\circ - \alpha$.

Now, let's focus on the median $AE$. We know that a median divides the side it is drawn to into two equal parts. This means that $AE=\frac{1}{2}c$. We can also use the fact that in a right triangle, the length of the median to the hypotenuse is half the length of the hypotenuse. This means that $AE=\frac{1}{2}c=\frac{1}{2}b$.

Using the angle bisector theorem, we can find the ratio $\frac{AD}{DC}=\frac{AB}{BC}=\frac{c}{a}$. Since $AD$ is the altitude, we can also use the Pythagorean theorem to find that $AD=\frac{1}{2}c\sqrt{2}$.

Now, we can use the fact that $AD$ is also a median to find that $AD=\frac{1}{2}c=\frac{1}{2}b$. This means that $AD=AE=\frac{1}{2}b$, which also implies that $DC=DE=\frac{1}{2}b$.

Using the Pythagorean theorem again, we can find that $AC=\sqrt{AD^2+DC^2}=\sqrt{\left(\frac{1}{2}b\right)^2+\left(\frac{1}{2}b