+115 Let PA and PC be tangents from P to a circle. Let B and D be points on the circle such that B, D, and P are collinear. Prove that AB*CD=BC*DA.
+129852 Note that m angle PAD = 1/2 m arc AD = m angle DBA
And
m angle PCD = 1/2 m arc DC = m angle DBC
And PA = PC....so....
sin PAD / PD = sin DBA / PD
sin PCD / PD = sin DBC/ PD
sin PDA/ PC = sin PAD / PD = sin DBA / PD
sin PDC / PC = sinPCD / PD = sin DBC / PD
PC / PD = sin PDA / sin DBA
PC / PD = sin PDC / sin DBC
So.....
sin PDA / sin DBA = sin PDC/ sin DBC
Note sin PDA = sin ADB because they are supplemental ...and..
sin PDC = sin BDC for the same reason....so....
sin ADB / sin DBA = sinBDC / sin DBC (A)
sin ADB / AB = sin DBA / DA
sinBDC / BC = sin DBC / DC
sin ADB / sin DBA = AB /DA
sin BDC / sin DBC = BC / DC
From (A) we must have, by substitution, that
AB / DA = BC / DC ⇒
AB * DC = BC * DA ⇒
AB * CD = BC * DA
The angle PAD is equal to the angle ABD, so the triangles PAD and PAB are similar (since the angle APB is common to both triangles).
That allows you to write down the ratios
\(\displaystyle \frac{AP}{PB} = \frac{PD}{AP}=\frac{DA}{AB}.\)
If you now do the same thing for the triangles the other side of the line PB, and compare the two sets of ratios, you should see that the result drops out in one line.
Tiggsy
