A more elegant way to prove an Euclidean geometry theorem

alternative-proofeuclidean-geometrygeometry

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?

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Best Answer

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 $\angle HJH'$.

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