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.

MIRB14  Mar 4, 2018

4+0 Answers


Note that  m angle PAD  =  1/2 m arc AD  =  m angle DBA


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




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




cool cool cool

CPhill  Mar 4, 2018

Thanks, but is there a way to not use trig?

MIRB14  Mar 4, 2018
edited by MIRB14  Mar 4, 2018

I'm sure there is....but....I'm not a good enough geometer to figure it out....



cool cool cool

CPhill  Mar 4, 2018

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.



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