Cyclic Quadrilaterals Advanced

Question

$ABCD$ is a cyclic quadrilateral; $x,y,z$ are the distances of $A$ from the lines $BD, BC, CD$ respectively. Prove that $$\frac{BD}x = \frac{BC}y + \frac{CD}z.$$

I guess this can be done by multiple application of cot rule. I found a really beautiful solution in a book.

Answer 1

Let $A', B', D'$ be the orthogonal projections of $A$ onto the lines $BD, BC, CD$ respectively, so $AA' = x, \, AB' = y, \, AD' = z$. Then the three points $A', B', D'$ are collinear (Simson's line).

1. Triangle $ABB'$ is similar to triangle $ADD'$. Therefore, $$\frac{AB}{AD}= \frac{AB'}{AD'}=\frac{y}{z}$$

2. Triangle $ADC$ is similar to triangle $AA'B'$. Therefore $$\frac{AC}{AD}= \frac{AB'}{AD'}=\frac{y}{x}$$

3. Apply Ptolemy's theorem to $ABCD$ $$BD \cdot AC = BC \cdot AD + CD \cdot AB$$ and plug $$AB = \frac{y}{z}\, AD, \,\, AC = \frac{y}{x} \, AD$$ so $$BD \cdot \frac{y}{x} \, AD = BC \cdot AD + CD \cdot \frac{y}{z}\, AD$$ which simplifies to $$\frac{BD}{x} = \frac{BC}{y} + \frac{CD}{z}$$

Yet another solution:

Let $A', B', D'$ be the orthogonal projections of $A$ onto the lines $BD, BC, CD$ respectively, so $AA' = x, \, AB' = y, \, AD' = z$.

1. Then $A', B', D'$ lie on the Simson line and let $H$ be the orthogonal projection of $A$ on the Simson line $B'D'$ and let its length be $h = AH$;

2. Triangle $AA'B'$ similar to triangle $ADC$ and moreover $AH = h$ and $AB' = y$ are corresponding similar altitudes. Hence $$\frac{BC}{y} = \frac{A'D'}{h};$$

3. Triangle $AA'D'$ similar to triangle $ABC$ and moreover $AH = h$ and $AD' = z$ are corresponding similar altitudes. Hence $$\frac{CD}{z} = \frac{A'B'}{h}$$

4. Triangle $AB'D'$ similar to triangle $ABD$ and moreover $AH = h$ and $AA' = x$ are corresponding similar altitudes. Hence $$\frac{BD}{x} = \frac{B'D'}{h}$$

5. Collinearity of $A', B' D'$ equivalent to $$B'D' = A'B' + A'D' $$ which means $$B'D' = \frac{BD}{x} \, h = A'B' + A'D'= \frac{CD}{z}\, h + \frac{BC}{y} \, h$$ and after cancelling out $h$ on both sides of the equation $$\frac{BD}{x} = \frac{BC}{y} + \frac{CD}{z}$$

Answer 2

That identity can be seen as a consequence of the properties of circle inversion. We have that $A$ lies on the circumcircle of $BCD$ (with radius $R$), hence by considering a circle inversion with respect to $\Gamma_A$, circle centered at $A$ through the circumcenter of $BCD$, we have that the images $B',C',D'$ of $B,C,D$ are collinear. Let $d$ be the distance of $A$ from the line through $B',C',D'$.

We clearly have $B'C'+C'D'=B'D'$ by collinearity, so $$\frac{B'C'}{d}+\frac{C'D'}{d} = \frac{B'D'}{d}$$

On the other hand, $ABC$ and $AC'B'$ are similar triangles, hence $$\frac{B'C'}{d}=\frac{BC}{y}$$ and the claim readily follows.