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# help plz

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help I need to find lengths

$A$, $B$, $C$, and $D$ are points on a circle, and segments $\overline{AC}$ and $\overline{BD}$ intersect at $P$, such that $AP=8$, $PC=2$, and $BD=5$. Find $BP$, given that $BP < DP.$ Sep 6, 2023

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To find $$BP$$, you can use Ptolemy's Theorem which applies to cyclic quadrilaterals. The theorem states that in such a quadrilateral, $$AC \times BD = AB \times CD + BC \times AD$$. First, $$AC = AP + PC = 10$$ and $$BD = 5$$. We can express $$AB$$ and $$AD$$ in terms of $$BP$$ and $$DP$$ since $$AB = BP + DP$$ and $$AD = BP - DP$$. Applying Ptolemy's theorem: $10 \times 5 = (BP + DP) \times 2 + 10 \times (BP - DP)$ $50 = 2BP + 2DP + 10BP - 10DP$ $50 = 12BP - 8DP$ $12.5 = 3BP - 2DP$ Since $$BP < DP$$, $$DP - BP > 0$$. Let $$x = DP - BP$$. Then $$DP = BP + x$$. Substitute into $$3BP = 12.5 + 2DP$$: $3BP = 12.5 + 2(BP + x)$ $3BP = 12.5 + 2BP + 2x$ $BP = 12.5 + 2x$ $BP - 2x = 12.5$ Also, $$x = DP - BP$$: $x = DP - (12.5 + 2x)$ $x = -12.5 - 2x$ $3x = -12.5$ $x = -\frac{12.5}{3}$ $x = -\frac{25}{6}$ Finally, $BP = 12.5 + 2x$ $BP = 12.5 - \frac{25}{3}$ $BP = \frac{12.5}{3}$ $BP = \frac{25}{6}$ So $$BP = \frac{25}{6}$$.

Sep 6, 2023
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Ptolemy has nothing to do with this problem. Also, 5 - 25/6 = 5/6 < 25/6 so clearly that cannot be BP.

plaintainmountain  Sep 6, 2023
edited by plaintainmountain  Sep 6, 2023
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Sep 6, 2023
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my bad you can solve it :D

Sep 7, 2023
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Let BP  = x   and DP =  5 - x

By the Intersecting Chords Theorem we  have

DP * PB  = AP * PC

(5 - x) * x =  8 * 2

5x - x^2  =  16       rearrange  as

x^2 - 5x + 16 =  0

We can see that this has no real solution for  x  because the  discriminant is  negative.....   Sep 7, 2023