help plz

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 $BP = \frac{25}{6}$.

about 2.9289 units

my bad you can solve it :D

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.....