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Given circle $(O)$ with chord $AB$, let $I$ be a point on any position of the chord $AB$ (except $A$ and $B$ themselves). Draw two more chords, $CD$ and $EF$ so that the chord $CE$ does not intersect the chore $AB$ of the circle. $CF$ and $DE$ intersect $AB$ at $M$ and $N$ respectively. Prove that: $\frac{AM\times IB}{IM}=\frac{BN\times IA}{IN}$ and $\frac{1}{IA}+\frac{1}{IN}=\frac{1}{IB}+\frac{1}{IM}$

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Attempt:

Obviously, with the case $IM=IN$, we can solve it by using the butterfly theorem, this theorem is quite popular. In this case however, $I$ is a point on any position of the chord $AB$.

I think the first problem might be related to the second problem as both equalities are quite similar to each other, I have attempted to use the Haruki lemma to prove that $AM\times IB=IM\times AG$ and $BN\times IA=IN\times BH$ but I'm stuck at this point, so proving $AG=BH$ would be neccessary if I'm following the right track (I draw two extra circles to prove the Haruki lemma as above).

Please note that I haven't learned about the symbol $\left(mod \pi \right)$, any solution should not have any relation to this (when I see some other difficult geometry problems been asked in my country, some people have used this symbol to answer, which I can't understand).

user061703
  • 3,877

2 Answers2

2

Using cross-ratios, $$ (AIMB) = (ADFB) = (ANIB) $$ or $$ \dfrac{AM}{IM}:\dfrac{AB}{IB} = \dfrac{AM}{DF}:\dfrac{AB}{DF} = \dfrac{AI}{NI}:\dfrac{AB}{NB}. $$ Hence, $$ \dfrac{AM\cdot IB}{IM} = \dfrac{AI\cdot NB}{NI}. $$

Instead of cross-ratios you may use triangle areas as well: $$ \dfrac{AM\cdot IB}{IM\cdot AB} = \dfrac{area[CAM]\cdot area[CIB]}{area[CIM]\cdot area[CAB]} = \dfrac{ (CA\cdot CM \cdot \sin\angle{ACM})\cdot (CI\cdot CB \cdot \sin\angle{ICB})}{ (CI\cdot CM \cdot \sin\angle{ICB})\cdot (CA\cdot CB \cdot \sin\angle{ACB})} = \dfrac{ \sin\angle{ACM})\cdot\sin\angle{ICB}}{ \sin\angle{ICM})\cdot\sin\angle{ACB}} = \dfrac{ \sin\angle{AEI})\cdot\sin\angle{NEB}}{ \sin\angle{NEI})\cdot\sin\angle{AEB}} = \dfrac{ (EA\cdot EI \cdot \sin\angle{AEI})\cdot (EN\cdot EB \cdot \sin\angle{NEB})}{ (EN\cdot EI \cdot \sin\angle{NEB})\cdot (EA\cdot EB \cdot \sin\angle{AEB})} = \dfrac{area[EAI]\cdot area[ENB]}{area[ENI]\cdot area[EAB]} = \dfrac{IA\cdot BN}{IN\cdot AB}. $$

The second statement is simply equivalent, $$ AM\cdot IB \cdot IN = IA \cdot NB \cdot IM $$ $$ (IA - IM) \cdot IB \cdot IN = IA \cdot (IB - IN) \cdot IM $$ $$ \dfrac1{IM} - \dfrac1{IA} = \dfrac1{IN} - \dfrac1{IB}. $$

G.Kós
  • 14,297
1

I think you are on the right track and you are more than halfway done. Here's my solution: Applying the Intersecting Chord Theorem repeatedly, we will have:

$AM\cdot MB=CM\cdot MF=IM\cdot MG$

$BN\cdot NA=DN\cdot EN=IN\cdot NH$

Therefore, $\dfrac{AM\cdot IB}{IM}=\dfrac{MG}{MB}\cdot IB=\dfrac{MG}{MB}\cdot (MB-MI)=MG-\dfrac{MG}{MB}\cdot MI=MG-AM=AG$

$\dfrac{BN\cdot IA}{IN}=\dfrac{NH}{NA}\cdot IA=\dfrac{NH}{NA}\cdot (NA-NI)=NH-\dfrac{NH}{NA}\cdot NI=NH-IN=BH$

Now we are going to show that $AG=BH$. Notice that $\angle{CFE}=\angle{CDE}=\angle{IDE}=\angle{IHE}$, hence $M,F,H,E$ are on the same circle.

Therefore, we have $FI\cdot IE=MI\cdot IH$ and $FI\cdot IE=AI\cdot AB$. $MI\cdot (IH-IB)=(AI-MI)\cdot IB\implies BH=\dfrac{AM\cdot IB}{MI}$

Similarly, we have $G,C,N,D$ on the same circle and $AG=\dfrac{AM\cdot IB}{MI}$.

Now we have proved $AG=BH$ and that finishes the proof.

The part b) is just simple calculation:

$\dfrac{1}{IA}+\dfrac{1}{IN}=\dfrac{1}{IB}+\dfrac{1}{IM} \iff AN\cdot IB\cdot IM=BM\cdot IA\cdot IN \iff (AI+IN)\cdot IB\cdot IM=(IM+IB)\cdot IA\cdot IN \iff \dfrac{IA\cdot IB}{IN}+IB=\dfrac{IA\cdot IB}{IM}+IA \iff GI+IB=IA+IH \iff GB=AH \iff AG=BH$

yifan dai
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