A more elegant way to prove an Euclidean geometry theorem

Question

Circle c (center $A$) and circle d (center $D$) is given such that the center of c lies on d. $DA$ intersects d at $C$ and $A$. $HI$ is a chord of c that goes through $C$. $EF$ is the radical axis of c and d that intersects $HI$ at $G$.

Prove that JG bisects $\widehat{HJI}$.

My proof currently is:

1. $AK\bot CK$ at K (Thales' theorem)
2. Thus $AK\bot HI$ at K
3. Since $HI$ is a chord of c $\Rightarrow HK=KI=1/2HI$
4. Applying intersecting chords theorem to both c and d we get $GC\cdot GK=GH\cdot GI=GE\cdot GF$
5. $\Rightarrow GH\cdot (CI-CG)=(GH+HC)(HK-HG)$
6. $\Leftrightarrow GH\cdot CI-GH\cdot CG=(GH+HC)((CI-CH)/2-HG)$
7. $\Leftrightarrow GH\cdot CI-GH\cdot CG=(GH+HC)((CI-CH)/2)-GC\cdot HG$
8. $\Leftrightarrow GH\cdot CI=(GH+HC)(CI-CH)/2$
9. $\Leftrightarrow 2GH\cdot CI=(GH+HC)(CI-CH)$
10. $\Leftrightarrow 2GH\cdot CI=GH\cdot CI-GH\cdot CH+CH\cdot CI-CH^2$
11. $\Leftrightarrow GH\cdot CI=(CI-GH-CH)\cdot CH$
12. $\Leftrightarrow GH\cdot CI=CH\cdot GI$
13. $\Leftrightarrow GH/GI=CH/CI$
14. Since $\widehat{CJG}$ is a right angle we can use the inverse angle bisector theorem to arrive at the result.

Evidently the algebra part is as painful as can be if not downright impossible if you don't know the target equation firsthand.

Questions:

1. Does this theorem have a name?
2. Is there a way to prove this that circumvents the algebraic manipulation?
3. Failing that, can the manipulation at least be made less cumbersome?

Answer 1

One can prove, equivalently, that $AC$ bisects $\angle HJH'$, where $H'$ is the other intersection of $IJ$ with $c$. From: $$ CH\cdot CI =CE^2=CA\cdot CJ $$ we get $$CH:CA=CJ:CI,$$ that is triangles $CHA$, $CJI$ are similar. It follows that $$ \angle CIJ=\angle HAC={1\over2}\angle HAH'. $$ Hence $AC$ bisects $\angle HAH'$ and also $\angle HJH'$.

Answer 2

Okay, there is a way to solve this problem quite quickly, but it uses cross ratio which some might consider to be too advanced or unnecessary.

Since by Thale's $CF\perp AF$ and $CE\perp EA$, $CE$ and $CF$ are the tangents from $C$ to circle $c$. Therefore $EHFI$ is a harmonic quadrilateral and $(EF;HI)=-1$. Then \begin{align*} (EF;HI)&\stackrel F=(G,FF\cap HI;H,I)\\ &=(GC;HI) \end{align*} since $CF$ is tangent to $c$. Then because $JC\perp JG$, a well-known theorem about cross ratios tells us that $JG$ bisects $\angle HJI$.

Answer 3

Here is an alternative way to proceed. The given situation is simplified, and/or reshaped equivalently in successive steps.

• $(1)$ First of all, the circle $\odot(d)$ is not needed, so we remove it together with its center from the picture, all we need from its reminiscence is that $CE,CF$ are tangents to the circle $\odot(A)$. (A circle is invoked via its center, if only one such circle is given in the picture.)
• $(2)$ The point $J$ is in the posted question tacitly $J=AC\cap EF$. Among the many points $E,F,G,J$ we keep only $J,E$, remove the other two. So we need to show that $JE$ is angle bisector of $\widehat{HJI}$.
• $(3)$ We construct the points $H_0,I_0$ as intersection points of the ray $CA$, with the circle $\odot(A)$. We fix their choice is that on the ray the points come in the order $C,H_0,J,A,I_0$, so $H_0$ comes first, as is also happens with $H$ on the ray $CHI$. Let $X$ be the intersection $X=H_0H\cap I_0I$. We take a closer look to the triangle $$ \Delta XH_0I_0\qquad\text{ with heights $I_0H$, $H_0I$, and $X?$ .} $$ Well, note that the third height in the list above has the property of bisecting the angle in its foot $?$ to the other feet $H_0,I_0$. So we have to show a simpler property.
• $(4)$ Putting all together, we have to show equivalently, building the picture which introduces the same points in a different order, namely $X,H_0,I_0;H,I,J;A,C,E$:

Proposition: Consider the triangle $\Delta XH_0I_0$ with heights $XJ$, $H_0I$, $I_0H$. Let $A$ be the mid point of $H_0I_0$, and $C$ the intersection $C=H_0AI_0\cap HI$. Let $E$ be an intersection point of the height $XJ$ and the circle $\odot(A)$ centered in $A$ with radius $AH_0=AI_0$. Then $CE\perp EA$.

For the proof, we have to show that $\Delta CEA$ has a right angle in $C$, which is equivalent to $EJ^2=JC\cdot JA$. (All segments have positive measure in this answer.) We know that $\Delta EH_0I_0$ has a right angle in $E$, so we know $EJ^2=JH_0\cdot JI_0$.

• $(5)$ At this point, $E$ can be also removed from the picture, we need to show equivalently without it just $JC\cdot JA=JH_0\cdot JI_0$.
• $(6)$ The theorems of Ceva and Menelaus applied in $\Delta XH_0I_0$ for the three heights as cevians, and for the "secant" $CHI$ show that $C,J$ are harmonic conjugates w.r.t. $H_0I_0$. This information is enough now to reduce the problem from 2D to 1D, we no longer need the points $X,H,I$, so remove them from the picture.
• $(7)$ Let us state explicitly what we have to show in one dimension.

Proposition: Consider two points $I_0,H_0$. Let $A$ be the mid point of the segment $I_0H_0$. On the ray $[I_0AH_0$ consider points $J\in [AH_0]$, and $C$ beyond $H_0$, so that $(C,J;H_0,I_0)$ is harmonic, i.e. $\displaystyle\frac{CH_0}{CI_0}=\frac{JH_0}{JI_0}$. (No signs again.) Then $JC\cdot JA=JH_0\cdot JI_0$.

The final picture deserves to be inserted in its own line:

We use a coordinate system with $I_0$ in $-1$, $A$ in $0$, $J$ in $x\in(0,1)$, $H_0$ in $+1$, and $C$ in $c>1$. The known "harmonic condition" is: $$ \frac {c-1}{c+1} =\frac{1-x}{1+x}\ . $$ And we want to show $(c-x)\cdot x=(1-x)(1+x)$. Both relations collapse to the same $cx=1$, so we are done.