Papillon - New Year Challenge

[Pedja]

Can you prove the following claim:

Let [imath]M[/imath] be the arbitrary point of a chord [imath]AB[/imath] of a circle different from [imath]A[/imath] and [imath]B[/imath] , through which two other chords [imath]CD[/imath] and [imath]EF[/imath] are drawn. [imath]CF[/imath] and [imath]ED[/imath] intersect chord [imath]AB[/imath] at [imath]P[/imath] and [imath]Q[/imath] correspondingly. Denote [imath]PM[/imath] by [imath]a[/imath] , [imath]MQ[/imath] by [imath]b[/imath] , [imath]AP[/imath] by [imath]c[/imath] and [imath]QB[/imath] by [imath]d[/imath] . Then , [imath]\frac{a+c}{b+d}=\frac{bc}{ad}[/imath] .

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P.S.
I will post my solution in two weeks.

 

[Pedja]

Proof.
Consider the following diagram:

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Since angles subtended on the same arc are equal we have:
[math]\angle P_1CP=\angle QEQ_2[/math][math]\angle PFP_2=\angle Q_1DQ[/math]Also, since vertical angles are equal we have:
[math]\angle PMP_1=\angle QMQ_1[/math][math]\angle P_2MP=\angle Q_2MQ[/math]Therefore,
[math]\triangle P_1CP \sim \triangle QEQ_2 \Longrightarrow \frac{a_1}{b_2}=\frac{CP}{EQ}[/math][math]\triangle PFP_2 \sim \triangle Q_1DQ \Longrightarrow \frac{a_2}{b_1}=\frac{PF}{QD}[/math][math]\triangle PMP_1 \sim \triangle QMQ_1 \Longrightarrow \frac{a}{b}=\frac{a_1}{b_1}[/math][math]\triangle P_2MP \sim \triangle Q_2MQ \Longrightarrow \frac{a}{b}=\frac{a_2}{b_2}[/math]Combining these four equalities we get:
[math]\frac{a^2}{b^2}=\frac{a_1}{b_1}\cdot \frac{a_2}{b_2}=\frac{a_1}{b_2}\cdot \frac{a_2}{b_1}=\frac{CP}{EQ}\cdot \frac{PF}{QD}[/math]According to intersecting chords theorem we have:
[math]\frac{CP}{EQ}\cdot \frac{PF}{QD}=\frac{c(a+b+d)}{d(b+a+c)}[/math]Hence,
[math]\frac{a^2}{b^2}=\frac{c(a+b+d)}{d(b+a+c)}[/math][math]\frac{b^2c(a+b+d)}{a^2d(b+a+c)}=1[/math][math]\frac{bc}{ad}=\frac{a(b+a+c)}{b(a+b+d)}[/math][math]\frac{bc}{ad}=\frac{ab+a(a+c)}{ab+b(b+d)}[/math][math]\frac{bc}{ad}=\frac{ab+bc+a(a+c)-bc}{ab+ad+b(b+d)-ad}[/math][math]\frac{bc}{ad}=\frac{b(a+c)+a(a+c)-bc}{a(b+d)+b(b+d)-ad}[/math][math]\frac{bc}{ad}=\frac{(a+b)(a+c)-bc}{(a+b)(b+d)-ad}[/math][math]bc(a+b)(b+d)-bcad=ad(a+b)(a+c)-bcad[/math][math]bc(a+b)(b+d)=ad(a+b)(a+c)[/math][math]\frac{a+c}{b+d}=\frac{bc}{ad}[/math]

[imath]\blacksquare[/imath]​