Show that $\frac{1}{a}=\frac{1}{b}+\frac{1}{c}$ in the triangle below

Question

In the figure, show that:$\dfrac{1}{c}=\dfrac{1}{a}+\dfrac{1}{b}$

I try: $AD$ and $BE$ are angle angle bissectors.

$\dfrac{x}{c} = \dfrac{a-x}{b} \implies x= \dfrac{ac}{b+c}$

$\dfrac{y}{c}= \dfrac{b-y}{a} \implies y =\dfrac{bc}{a+c} $

$CD =a-\dfrac{ac}{b+c} \implies CD = \dfrac{ab}{b+c}$

$\triangle ADC \sim \triangle BAC:\dfrac{\dfrac{ab}{b+c}}{b}=\dfrac{b}{a} \implies a^2=b^2+bc(I)$

$\triangle ABE \sim \triangle ACB:\dfrac{\dfrac{bc}{a+c}}{c}=\dfrac{c}{b} \implies b^2=c^2+ac(II)$

$(I):(II):a^2 = c^2+ac+bc \implies \dfrac{a^2-c^2}{abc}=\dfrac{1}{a}+\dfrac{1}{b}$

How to finish? I need a resolution by plane geometry if possible.

Answer 1

You just need to prove that $\;a^2-c^2=ab\,.$

Let $\,BH\,$ be the perpendicular segment from the vertex $\,B\,$ to the opposite side $\,AC\,$ and let $\,F\,$ be the point which belongs to the extension of the segment $\,CH\,$ such that $\,FH\cong HA\,.$

The triangle $\,BFA\,$ is isosceles and it results that

$\angle BFA=\angle BAF=\pi-4\theta\;\;,$

moreover ,

$\angle FBC=\pi-\angle BCF-\angle BFA=\pi-\theta-(\pi-4\theta)=3\theta\,.$

Since $\,7\theta=\pi\,,\,$ it follows that $\,\angle BFA\cong\angle FBC\,,\,$ consequently also the triangle $\,BFC\,$ is isosceles, therefore
$FC=a\,$ and $\,HA=\dfrac{a-b}2\,.$

Furthermore, by applying Pythagoras' theorem to the triangles $\,BHC\,$ and $\,BHA\,,\,$ it results that

$BH^2=a^2-\left(b+\dfrac{a-b}2\right)^2=a^2-\left(\dfrac{a+b}2\right)^2\;\;,$

$BH^2=c^2-\left(\dfrac{a-b}2\right)^2\;\;,$

hence ,

$a^2-\left(\dfrac{a+b}2\right)^2=c^2-\left(\dfrac{a-b}2\right)^2\;\;,$

$a^2-c^2=\left(\dfrac{a+b}2\right)^2-\left(\dfrac{a-b}2\right)^2\;\;,$

therefore ,

$a^2-c^2=ab\,.$

Answer 2

You can simply do it this way:

We have $\frac{1}{a} + \frac{1}{b}= \frac{1}{2R}\left(\frac{1}{\sin(4\theta)}+\frac{1}{\sin(2\theta)}\right)$

Or, $\frac{1}{a}+\frac{1}{b} = \frac{1}{2R} \left(\frac{\sin(2\theta)+\sin(4\theta)}{\sin(4\theta)\sin(2\theta)}\right)$

As, $\sin A + \sin B = 2\sin(\frac{A+B}{2})\cos(\frac{A-B}{2})$, We can write it as

$\frac{1}{a} + \frac{1}{b} = \frac{1}{2R}\left(\frac{2\sin(3\theta)\cos(\theta)}{\sin(4\theta)\sin(2\theta)}\right)$

As $\theta + 2\theta + 4\theta = \pi$, we can cancel $\sin(3\theta)$ and $\sin(4\theta)$ and we are left with $\frac{1}{a}+\frac{1}{b} = \frac{1}{2R\sin(\theta)} = \frac{1}{c}$

Answer 3

Here is an another approach:

In the figure, $D$ is constructed such that $\angle DBC=\angle DCB=2\theta$. $DB$ and $CA$ are then extended to meet at $E$.

Since $ \angle ABC=\angle BCD=2\theta$, $AB//CD$

Furthermore note that $\angle BDC=3\theta$ and $\angle BEA=\theta.$

Using the facts that $\Delta EBA, \Delta EBC, \Delta ECD$ and $\Delta BCD$ are isosceles triangles, we can deduce that $EB=BC=a$, $BD=CD=b$ and hence $ED=a+b$

Since $\Delta EBA \sim \Delta EDC,$ we have $\frac{DC}{AB}=\frac{ED}{EB}$ and hence $$\frac{b}{c}=\frac{a+b}{a} $$

$$\implies \frac{1}{c}=\frac{a+b}{ab}$$ $$\implies \frac{1}{c}=\frac{1}{a}+\frac{1}{b}$$