Pedja
New member
- Joined
- Jul 8, 2021
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- 35
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] .
P.S.
I will post my solution in two weeks.
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] .
P.S.
I will post my solution in two weeks.