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The circles $O_1$ and $O_2$ do not intersect and do not have the same radius.

$\overline{AB}$ and $\overline{CD}$ are tangent lines of circles $O_1$ and $O_2$, with $A$ and $C$ on the circumference of $O_1$, $B$ and $D$ on the circumference of $O_2$.

Let $\overline{EF}$ be the third tangent line of both circles, with $E$ on the circumference of $O_1$, and $F$ on the circumference of $O_2$,

Extend $\overline{EF}$ to intersect $\overline{AB}$ at $G$ and $\overline{CD}$ at $H$.

Prove that $\overline{GE}$ = $\overline{FH}$

I drew it like this:

enter image description here

I got $AB = CD$ but couldn't find any relations.

Sorry for my bad grammar.

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  • $\begingroup$ What exactly is the question? $\endgroup$ – QuIcKmAtHs Jan 21 '18 at 3:45
  • $\begingroup$ Prove that GE = FH. $\endgroup$ – Khunpol Jermsiri Jan 21 '18 at 3:46
  • $\begingroup$ E and G are labelled wrongly $\endgroup$ – QuIcKmAtHs Jan 21 '18 at 3:58
  • $\begingroup$ oh sorry. i'll fix that $\endgroup$ – Khunpol Jermsiri Jan 21 '18 at 3:59
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$$GE=AE=AB-BE=CD-EF=CH+HD-(EG+GF)=$$ $$=HG+HF-EG-GF=(HG-GF)+HF-EG=2HF-EG$$ and we are done!

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Tangents from a point to a circle are equal, so $GE=GA$, $GB=GF$, $HD=HF$ and $HC=HE\,$. Then:

$$\require{cancel} GE+GF = GA+GB = AB = CD = HD+HC = HF+HE \\ \implies \quad GE+GF=HF+HE \quad \iff \quad 2 GE + \cancel{EF} = 2 HF+ \cancel{FE} $$

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