1. ## really hard! need help plz!

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

2. I attempted to solve d problem and I think I got it for the case when AB and CD are not equal.

Construction Steps:
(1) Draw perpendicular bisector of AD. Mark any arbitrary point M on it.
(2) Draw perpendicular bisector of BC. Mark N on it such that NB=AM. N and M are on same side of the line.
(3) Draw perpendicular bisector of MN. The point of intersection of this bisector with the original line gives P.

3. 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)

$x^2 + dx = x^2 + bx + cx + bc$

which results in

$x = \frac{bc}{d-b-c}$

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?

4. Originally Posted by aidan
$x = \frac{bc}{d-b-c}$

Are you asking how to construct a solution graphically?
Is that the question?
I think the question is to find the point graphically(using geometry). Also, what if
$d=b+c$

PS: I solved for the case d=b+c in next post.

5. the solution that I posted above was valid when AB is not equal to CD. This solution is valid when AB=CD.

Construction Steps:
(1)Draw perpendicular bisector of BC. Mark any arbitrary point "O" on it.

Let OA=R
and OB=r

(2) Draw a right angled triangle LMN, right angled at N such that MN=R and LN=r.

(3) Draw a isosceles right angled triangle LMK, right angled at K.

let MK=t

(4) Draw a circle with "O" as centre and radius=t. The points of intersection of the circle with the given line give two possible positions of P.