# How do I prove that $\left( 1-\frac{a}{b} \right)\left( 1+\frac{c}{d} \right)=4$?

The challenge is to prove $$\left(1-\dfrac{a}{b}\right)\left(1+\dfrac{c}{d}\right)=4.$$

Apart from the Pythagorean theorem, the tangent and secant theorem, and the cosine theorem, I could not invent anything. The calculations here turn out to be very cumbersome and I'm not sure that this is how it should be initially solved.

I had an idea to start from the similarity in the figure, there is a lemma about a right angle on a chord, there is a property of a tangent and a secant...But all this again will lead to cumbersome calculations.

• It might be useful to rearrange the equation as $$bc = a(c+d) + 3bd$$ which possibly could be deduced by a clever disection argument. But I dont see an easy one. Commented Jul 1, 2021 at 1:58

While we wait for an easier method, my answer is based on system of equations using Pythagoras and solving them by hand.

We will use $$CE = CF = e, CT = t$$ and $$c + d = x$$

By Tangent-secant theorem, we have $$\ CF^2 = CT \cdot CD \implies e^2 = t (t+x)$$ ...$$(i)$$

$$AC^2 = AH^2 + CH^2 \implies (a+e)^2 = a^2 - d^2 + (x + t - d)^2$$

$$e^2 + 2 a e = (x+t)^2 - 2 d (t+x)$$

Substituting $$e^2$$ from $$(i), 2ae = x^2 + tx - 2d(t+x)$$ ...$$(ii)$$

Now using similarity of $$\triangle ADH$$ and $$\triangle BDG$$, $$DG = \dfrac{bd}{a}$$

$$BC^2 = BG^2 + CG^2$$
$$\implies (b+e)^2 = b^2 - \left(\dfrac{bd}{a}\right)^2 + \left(t + x + \dfrac{bd}{a}\right)^2$$

$$2be = x^2 + tx + \dfrac{2bd}{a} (t+x)$$ ...$$(iii)$$

By $$(iii) - (ii), (b-a)e = \dfrac{(a+b)d}{a} (t+x)$$ ...$$(iv)$$

By $$(ii) \times b + (iii) \times a \ , \ 4abe = x (a + b) (t + x)$$

Using $$(iv), x = \dfrac{4 b d}{(b-a)}$$

As $$x = c + d, \ \dfrac{(c+d) (b-a)}{bd} = 4$$

$$\left(1 - \dfrac{a}{b}\right) \left(1 + \dfrac{c}{d}\right) = 4$$

It seems like the form of the target relation is trying to tell us something deeper about the configuration that I'm just not seeing. Nevertheless, here's a solution that amounts to a few direct calculations followed by a tedious change of parameters.

With different notation: Let $$\triangle ABC$$ (with $$\angle A\geq \angle B$$) have conventional sides $$a$$, $$b$$, $$c$$, which are touched by the incircle at points $$D$$, $$E$$, $$F$$. Let the pairs tangent segments from the vertices have lengths $$a'$$, $$b'$$, $$c'$$. Let cevian $$\overline{CF}$$ meet the incircle at $$U$$; and let the perpendicular from $$A$$ meet that cevian at $$V$$. Finally, let the perpendicular from $$C$$ meet side $$c$$ at $$F'$$; by Thales' Theorem, $$V$$ and $$F'$$ lie on the semicircle with diameter $$\overline{AC}$$.

Defining $$t:=|CU| \qquad u:=|UV| \qquad v:=|VF| \qquad f:=|CF|=t+u+v$$ the goal is to show $$\left(1-\frac{a'}{b'}\right)\left(1+\frac{u}{v}\right)=4 \tag{\star}$$

We can write $$1+\frac{u}{v}=\frac{u+v}{v}=\frac{f-t}{v}=\frac{f^2-ft}{fv} \tag1$$ Multiplying-through by $$f$$ gives us values we can calculate as follows: \begin{align} f^2 &= a'^2+b^2-2a'b\cos A &\text{(Law of Cosines)} \tag2\\ ft &= c'^2 &\text{(Sec-Tan Thm wrt incircle)} \tag3\\ fv &= |FA||FF'|=a'(a'-b\cos A) &\text{(Sec-Sec Thm wrt semicircle AC)} \tag4 \end{align}

Using the Law of Cosines $$\cos A = \frac{1}{2bc}(-a^2+b^2+c^2) \tag5$$ and rewriting $$a=b'+c'$$, etc, we find (after a round of symbol-crunching) $$f^2 - ft = \frac{4a'b'c'}{a'+b'} \qquad fv = \frac{a'c'(b'-a')}{a'+b'}\qquad\stackrel{(1)}{\to}\qquad 1+\frac{u}{v}=\frac{4b'}{b'-a'} \tag6$$ from which $$(\star)$$ follows. $$\square$$

So what I want to say is that I found what could be said to be the "best" way to solve this problem. First, consider the following drawing:

All colored $$\triangle$$ are pairwise similar to each other, $$\left(r_{a} r=a b\right)$$.

Yellow $$\sim$$ blue $$\Rightarrow \dfrac{b-a}{2 r}=\dfrac{B_{2} K_{2}}{c+d}$$.

Blue $$\sim$$ red $$\Rightarrow \dfrac{B_{1} K_{1}}{d}=\dfrac{2 r}{a}$$.

Blue $$\sim$$ green $$\Rightarrow \dfrac{B_{2} K_{2}}{B_{1} K_{1}}=\dfrac{r_{a}}{r}$$.

We get:

$$(b-a)(c+d)=2 B_{2} K_{2} \cdot r=2 B_{1} K_{1} \cdot r_{a}=4 r_{a} r \frac{d}{a}=4 b d$$