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)
$\displaystyle x^2 + dx = x^2 + bx + cx + bc $
which results in
$\displaystyle x = \frac{bc}{d-b-c} $
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?