How to prove $\measuredangle FIB +\measuredangle IBC - \measuredangle FCB = 90^{\circ}$?

Question

In triangle ABC point $G$ is the point of intersection of the line $AI$ with circumcircle of triangle ABC, where $I$ is the center of the circle inscribed. Let point $I'$ be orthogonal projection point $I$ on the line $BC$ and a point $F$ be the point of intersection of the line $GI'$ with circumcircle of triangle ABC. How to prove $\measuredangle FIB +\measuredangle IBC - \measuredangle FCB = 90^o$? Is there a simple way?

Answer 1

Well, this problem is not that difficult. First I'll make some changes such that it will be easier for me to write.$$\measuredangle BAC=\alpha$$ $$\measuredangle ABC=\beta$$ $$\measuredangle ACB=\gamma$$ $$\measuredangle AGF=x$$ Now, observe that $$\measuredangle FIB=180^{\circ}-(\measuredangle IFB+\measuredangle FBI)$$ Since $\measuredangle BFI=\measuredangle BAO=90^o-\gamma$ because look the same arc (1) and $\measuredangle FBI=\measuredangle \frac{\beta}{2}+x$, we get:$$\measuredangle FIB=180^{\circ}-(90^{\circ}-\gamma+x+\frac{\beta}{2})$$Now, $\measuredangle IBC=\frac{\beta}{2}$ and $\measuredangle FCB=\gamma-\measuredangle FCA=\gamma-x$ since $\measuredangle FCA=\measuredangle FGA$. Now we evaluate what is asked to shown: $$\measuredangle FIB+\measuredangle IBC-\measuredangle FCB=180^{\circ}-(90^{\circ}-\gamma+x+\frac{\beta}{2})+(\frac{\beta}{2})+(\gamma-x)=90^{\circ}$$

And it is done!

(1) The prove that they look at the same arc:

Lemma 1: This is a well known lemma, the reflexion of the orthocenter of $ABC$ is $P$ when we reflect the orthocenter throw $BC$, and when we reflect the orthocenter through the mid-point of $BC$ it will be $P'$ that is also the reflection of $P$ through $GO$. Moreover $P'OA$ are colinear.

The proof of lemma 1 is very easy, simply angle chasing, so to not make the solution larger I'll let you to prove it.

Lets now observe that, if $\angle I'IG=\measuredangle HAG$ is equal to $\measuredangle IFG$ then $FI$ and $P'$ will be on the same line, since $GA$ is the bisector of $PAP'$ from lemma 1. Now to prove that $\measuredangle I'IG=\angle IFG$ is sufficient to prove that triangles $II'G$ is similar to $FIG$. But they have in commom the angle $\measuredangle IGF$ so we only need to prove that: $$\frac{IG}{FG}=\frac{I'G}{IG}$$ $$IG^2=FG\cdot I'G$$

Now observe that the power of the point $I'$ is $I'C\cdot I'B=I'G\cdot I'F$ now $I'F=FG-I'G$ so $I'C\cdot I'B=I'G\cdot (FG-I'G)=I'G\cdot FG-I'G^2$. Form here $$I'C\cdot I'B+I'G^2=I'G\cdot FG=IG^2$$ Now one of the properties of the incenter is that $IG=GB$ (angle chasing). Hence we have $$IG^2=GB^2=I'C\cdot I'B+I'G^2$$ By cosine theorem $$I'G^2=BI'^2+GB^2-2BI'GB\cos (\measuredangle GBC)$$. Now evaluating $$GB^2=I'C\cdot I'B+I'G^2=I'C\cdot I'B+BI'^2+GB^2-2BI'GB\cos (\measuredangle GBC)$$ You will get in some point $$BI'+I'C=2BG\cos(\measuredangle GBC)=BC$$ Obviously true. Hence we are done!

I noticed that there is another proof easier to $GB^2=I'G\cdot FG$ since triangles $FGB$ is similar to $BI'G$ since $\measuredangle I'BG=\measuredangle GCB=\measuredangle BFG$, hence $$\frac{GB}{FG}=\frac{I'G}{GB}$$ And done :)