# Angle Bisectors Cross Circumcircle

### Problem

Let $I$ be the incenter of $\Delta ABC$ and points $A', B', C'$ the intersections of the extended angle bisectors $AI,BI,CI$ with the circumcircle of the triangle.

Prove that

1. $\displaystyle\frac{IA'\cdot IC'}{IB}=R$ and

2. $\displaystyle\frac{IA\cdot IC}{IB'}=2r,$

where $R$ is the circumradius, $r$ the inradius, of $\Delta ABC.$

### Hint

I do not know if you can do that without using some trigonometric identities in a triangle. Also, having intersecting chords in a circle brings to mind a certain well-known theorem.

### Solution

According to the Intersecting Chords theorem, say, $IA\cdot IA' = IB\cdot IB' = IC\cdot IC',$ implying

$\displaystyle\frac{IA'\cdot IC'}{IB}=\frac{IB'\cdot IC'}{IA}.$

Similarly,

$\displaystyle\frac{IA'\cdot IC'}{IB}=\frac{IB'\cdot IC'}{IA}=\frac{IA'\cdot IB'}{IC}$

and

$\displaystyle\frac{IA\cdot IC}{IB'}=\frac{IB\cdot IC}{IA'}=\frac{IA\cdot IB}{IC'}.$

It follows that what we have to prove is equivalent to

$\displaystyle\frac{(IA'\cdot IB'\cdot IC')^2}{IA\cdot IB\cdot IC}=R^3$

and

$\displaystyle\frac{(IA\cdot IB\cdot IC)^2}{IA'\cdot IB'\cdot IC'}=(2r)^3.$

Let $\angle A=2\alpha,$ $\angle B=2\beta,$ and $\angle C=2\gamma.$ By the Sine Law,

$\displaystyle\frac{IA}{AC}=\frac{\mbox{sin}\gamma}{\mbox{sin}(\pi -\alpha -\gamma)}=\frac{\mbox{sin}\gamma}{\mbox{cos}\beta}.$

In other words, $\displaystyle IA=b\cdot\frac{\mbox{sin}\gamma}{\mbox{cos}\beta}.$

Similarly, $\displaystyle IB=c\cdot\frac{\mbox{sin}\alpha}{\mbox{cos}\gamma}$ and $\displaystyle IC=b\cdot\frac{\mbox{sin}\beta}{\mbox{cos}\alpha}.$

Now note that $\angle IAB'=\alpha+\beta=\angle AIB',$ making $\Delta AB'I$ isosceles so that $AB'=IB'.$ It follows that $\displaystyle\frac{IB'}{AB}=\frac{AB'}{AB}=\frac{\mbox{sin}\beta}{\mbox{sin}2\gamma}.$ In other words, $\displaystyle IB'=c\frac{\mbox{sin}\beta}{\mbox{sin}2\gamma}.$ Similarly, $\displaystyle IC'=a\frac{\mbox{sin}\gamma}{\mbox{sin}2\alpha}$ and $\displaystyle IA'=b\frac{\mbox{sin}\alpha}{\mbox{sin}2\beta}.$

Multiplying what we got so far,

$\displaystyle IA\cdot IB\cdot IC=abc\frac{\mbox{sin}\alpha\cdot\mbox{sin}\beta\cdot\mbox{sin}\gamma}{\mbox{cos}\alpha\cdot\mbox{cos}\beta\cdot\mbox{cos}\gamma}$

and

$\displaystyle IA'\cdot IB'\cdot IC' =abc\frac{\mbox{sin}\alpha\cdot\mbox{sin}\beta\cdot\mbox{sin}\gamma}{\mbox{sin}2\alpha\cdot\mbox{sin}2\beta\cdot\mbox{sin}2\gamma}.$

From the double argument formulas,

$\displaystyle \frac{\mbox{sin}^{2}\alpha}{\mbox{sin}^{2}2\alpha}\cdot\frac{\mbox{cos}\alpha}{\mbox{sin}\alpha}=\frac{1}{2\mbox{sin}2\alpha}$

and

$\displaystyle \frac{\mbox{sin}2\alpha}{\mbox{sin}\alpha}\cdot\frac{\mbox{sin}^{2}\alpha}{\mbox{cos}^{2}\alpha}=\frac{2\mbox{sin}^{2}\alpha}{\mbox{cos}\alpha}$

all that remains is to check that

$\displaystyle R^{3}=\frac{abc}{8\mbox{sin}2\alpha\cdot\mbox{sin}2\beta\cdot\mbox{sin}2\gamma}$

and

$\displaystyle 8r^{3}=\frac{8\space {abc}\mbox{sin}^{2}\alpha\cdot\mbox{sin}^{2}\beta\cdot\mbox{sin}^{2}\gamma}{\mbox{cos}\alpha\cdot\mbox{cos}\beta\cdot\mbox{cos}\gamma}.$

But the first identity follows from the law of sines:

$\displaystyle 2R=\frac{a}{\mbox{sin}\alpha}=\frac{b}{\mbox{sin}\beta}=\frac{c}{\mbox{sin}\gamma}$

while the second one is a consequence of

$\displaystyle r=IA\cdot\mbox{sin}\alpha=b\frac{\mbox{sin}\alpha\cdot\mbox{sin}\gamma}{\mbox{cos}\beta}=a\frac{\mbox{sin}\beta\cdot\mbox{sin}\gamma}{\mbox{cos}\alpha}=c\frac{\mbox{sin}\alpha\cdot\mbox{sin}\beta}{\mbox{cos}\gamma}.$

### Acknowledgment

This is problem 11.7 from V. V. Prasolov's Problems in Planimetry (v 1, Nauka, Moscow, 1986 (Russian).)

### Angle Bisector

• Angle Bisector
• Angle Bisector Theorem
• Angle Bisectors in Ellipse
• Angle Bisectors in Ellipse II
• Angle Bisector in Equilateral Trapezoid
• Angle Bisector in Rectangle
• Property of Angle Bisectors
• Property of Angle Bisectors II
• A Property of Angle Bisectors III
• External Angle Bisectors
• Projections on Internal and External Angle Bisectors
• Angle Bisectors On Circumcircle
• Angle Bisectors in a Quadrilateral - Cyclic and Otherwise
• Problem: Angle Bisectors in a Quadrilateral
• Triangle From Angle Bisectors
• Property of Internal Angle Bisector - Hubert Shutrick's PWW
• For Equality Choose Angle Bisector