Originally Posted by

**calculus_jy** I am desperate to solve this question :

Given 4 points A,B,C,D on a straight line, show how to construct a point P on the line such that PA.PD=PB.PC

The theorems that can be applied concerns intersecting chords and tangents

Not sure what is intended by: PA.PD=PB.PC

However,

If the object is to find some point P,

such that the distance PA times distance PD

is equal to

the distance PB times distance PC, then

Assign x to the distance PA.

Assign b to the distance AB

assign c to the distance AC

assign d to the distance AD

Code:

..P....................A.......B............C.......D...
..|<--------x--------->|<--b-->|............|.......|...
..|<--------x--------->|<--------------c--->|.......|...
..|<--------x--------->|<-----------------------d-->|...

x(x+d) = (x+b)(x+c)

which results in

for some random points along the x-axis

A = 37.97089586895891

B = 41.26128066028468

C = 46.5861084072385

D = 54.18555800267495

From the above

x= 6.57854211961539

P = A-x = 31.39235374934351

The distances:

PA = 6.57854211961539 = x

PB = 9.86892691094117 = x+b

PC = 15.1937546578949 = x+c

PD = 22.7932042533314 = x+d

PA*PD= 149.9460542215376

PB*PC= 149.9460542215376

There also exists a point P to the right of the A above.

The problem has a solution.

Are you asking how to construct a solution graphically?

Is that the question?