+817 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 $BP < DP.$
+122 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 \( BP = \frac{25}{6} \).
+170 Ptolemy has nothing to do with this problem. Also, 5 - 25/6 = 5/6 < 25/6 so clearly that cannot be BP.
