# Prove $XX’+ZZ’=YY’$

Let the feet of angle bisectors of triangle $$ABC$$ are $$X,Y$$ and $$Z$$ . The circumcircle of triangle $$XYZ$$ cuts off three segment from lines $$AB,BC$$and $$CA$$, let it be $$XX’, YY’, ZZ’$$. Prove that at least one combination exists such that sum of two segment's length is equal to third segment’s length.

WLOG, I assumed

$$YY’\geq XX’\geq ZZ’$$, So we have to prove $$YY’=XX’+ZZ’$$

I tried it using coordinate geometry assuming $$A,B,C$$ to be $$(x_i,y_i)$$ for $$i=1,2,3$$ . Then we can find coordinates of $$X,Y,Z$$ and hence the equation of circumcircle of $$\Delta XYZ$$. And then intercept with three sides. But it will consume a whole to day If a human follows this method without computer help.

• Maybe I did something wrong, by the intersecting secant theorem we have $$CY\cdot YY'= CX \cdot XX'$$ and similar relation for the other two pairs of secant. Assume the triangle is equilater, then perpendicular bisector are heights and so the are the height of the triangle $\triangle CXY$, and so such triangle is isosceles, thus $CX=CY$ and so $$XX'=YY'$$. By repeating the process for $YY'$ and $ZZ'$ you obtain $$XX'=YY'=ZZ'$$ and the theorem is true only they are all zero. I can't spot my mistake. Is there some condition on the triangle? Commented Mar 3 at 12:02
• @Marco by intersecting secant theorem, $$CY\cdot CY’= CX\cdot CX’$$ Commented Mar 3 at 12:46
• Right, but if $CX=CY$ (that happen when the triangle is equilateral) then you have $$CX(CX+XX')=CY(CY+YY') \rightarrow XX'=YY'$$ Commented Mar 3 at 13:08
• @Marco yes all are zero in equilateral triangle. Since the given circle will be incircle in that case. Commented Mar 3 at 13:12
• What is the source? Most euclidean geometry questions like this are from a contest.
– D S
Commented Mar 3 at 15:12

For the sake of simplicity, let's assume $$AB=c, AC=b, AB=c, XX'=x, YY'=y,$$ and $$ZZ'=z$$. Just by using the properties of the power of a point and angle bisectors, we will have:

$$\frac{ac}{b+c}(\frac{ac}{b+c}-x)=\frac{ca}{a+b}(\frac{ca}{a+b}+z), \\ \frac{bc}{a+b}(\frac{bc}{a+b}-z)=\frac{bc}{a+c}(\frac{bc}{a+c}-y), \\ \frac{ab}{a+c}(\frac{ab}{a+c}+y)=\frac{ab}{b+c}(\frac{ab}{b+c}+x),$$

or equivalently,

$$\frac{1}{b+c}(\frac{ac}{b+c}-x)=\frac{1}{a+b}(\frac{ca}{a+b}+z), \\ \frac{1}{a+b}(\frac{bc}{a+b}-z)=\frac{1}{a+c}(\frac{bc}{a+c}-y), \\ \frac{1}{a+c}(\frac{ab}{a+c}+y)=\frac{1}{b+c}(\frac{ab}{b+c}+x).$$

By solving this $$3-$$variable system of equations, we will get:

$$2x=\frac{a(c-b)}{b+c}+\frac{b(b+c)}{a+c}-\frac{c(b+c)}{a+b}, \\2z=\frac{c(b-a)}{a+b}+\frac{a(a+b)}{b+c}-\frac{b(a+b)}{a+c}, \\ 2y=\frac{b(c-a)}{a+c}+\frac{a(a+c)}{b+c}-\frac{c(a+c)}{a+b}.$$

Now, it's almost obvious that we have $$2x+2z=2y.$$

We are done.

Note $$1$$: To clarify the initial relations, we have used the fact that, for example, $$BX=\frac{ac}{b+c}$$ and $$BX(BX-XX')=BZ(BZ+ZZ').$$

Note $$2$$: The $$3-$$variable system of linear equations can be easily solved by hand.

• It seems this computational approach is unavoidable, as the best I can show is that AX',BY',CZ' are concurrent.
– D S
Commented Mar 3 at 15:24
• @DS Usually, these contest problems and their claims don't come out of nowhere. So, it is likely for the official solution to be non-computational. However, if this is a contest and the OP has to get the scores, the best way is not always the shortest or the most elegant one. If this is just a challenge or practice, the OP can (or should !) spend a lot of time not only to find the nicer solution but also to gain the some geometric intuition through the thinking process. I hope you (or someone else) will come up with an elegant approach! Commented Mar 3 at 15:41

I will outline an another approach that is useful to remember for Euclidean/Olympiad geometry problems. A simple application of Theorem of Sines gives us: $$XX' = 2r\sin\left(\frac\alpha 2+\gamma\right)=2r\sin\left(\frac\alpha 2+\beta\right)=r\left(\sin\left(\frac\alpha 2+\gamma\right)+\sin\left(\frac\alpha 2+\beta\right)\right) =$$ $$=2r\sin\frac{\pi}{2}\cos\frac{\beta-\gamma}{2} = 2r\cos\frac{\beta-\gamma}{2}.$$

From here, it suffices to prove these $$\cos\frac{\beta-\gamma}{2}, \cos\frac{\gamma-\alpha}{2}, \cos\frac{\alpha-\beta}{2}$$ satisfy your condition. Then you can brute force this however way you want - I personally would start out by finding: $$\sin\frac{\alpha}{2} = \sqrt{\frac{1-\cos\alpha}{2}} = \sqrt{\dfrac{(a-b+c)(a+b-c)}{4bc}} = \sqrt{\frac{(p-b)(p-c)}{bc}}$$ and then you get: $$\cos\frac{\alpha}{2} = \sqrt{\dfrac{p(p-a)}{bc}}$$ and so on.