BA^2 + CD^2 = 16
PA = 1.25
PC = 0.75
AB? CD?
Thank you!
What have you tried?
Do you know the 'intersecting secant theorem'?
Look here:
Intersecting Secant Theorem - Math Open Reference
Along the theorem:
$\displaystyle PA\cdot PB = PC \cdot PD$
But we can write $\displaystyle PB=PA+AB$ and $\displaystyle PD=PC+DC$ therefore:
$\displaystyle PA\cdot (PA+AB)=PC \cdot (PC+DC)$
$\displaystyle \Leftrightarrow (PA)^2+PA\cdot AB=(PC)^2+PC\cdot DC$
$\displaystyle \Leftrightarrow DC=\frac{(PA)^2+PA\cdot AB - (PC)^2}{PC}$ (1)
You have given $\displaystyle PA=1,25$ and $\displaystyle PC=0,75$ so you can substitute them in (1). You have also given:
$\displaystyle (AB)^2+(DC)^2=16$ (2)
If we susbtitute (1) in (2) we get:
$\displaystyle (AB)^2+\left[\frac{(PA)^2+PA\cdot AB - (PC)^2}{PC}\right]^2=16$
If you have substituted the given values for PC and PA into this equation then you will have a quadratic equation in one variable AB which you can solve (you will get two solutions but offcourse you have to reject the negative one because the lenght of a side can't be negative).
You're welcome! Did you find AB and CD?
If you want a proof then you can take a look here:
http://www.mathforamerica.org/c/docu...e=DLFE-112.pdf