"Given four points, A, B, C, D in order on a straight line construct a point P on BC such that PA.PB = PC.PD"
I assume the end result is to have two right angled triangles AXP with X perpendicular to B and angle AXP a rt angle and PYD with Y perpendicular to C and < PYD a rt angle , PX = PY (being the mean proportionals required) but I cannot see how to get there! What am I missing? Any ideas?
On
This is an easier method.
Step-1 Move the line segment BC vertically upward a number of units to B’C’.
Step-2 Let AC’ and DB’ intersect at X.
Step-3 Draw XT, the perpendicular from X to AD.
In figure (1), $\frac {XD}{XB’} = \frac {XA}{XC’}$
In (2), $\frac {XD}{XB’} = \frac {TD}{TB}$
In (3), $\frac {XA}{XC’} = \frac {TA}{TC}$
∴ $\frac {TD}{TB} = \frac {TA}{TC}$ and the required result follows.
The construction is not that direct – and I don’t like it either. Anyway, here it goes:-
Let $AB = x$, $BC = y$ and $CD = z$ be the given lengths of the straight line $ABCD$.
We further let $P$ be at a distance $t$ from $B$.
$PA.PB = PC.PD$ means $(t + x)t = (y – t)(y – t + z)$
Simplifying and re-arranging, we have $t = \frac {y + z}{x + y + z + y}y$ [Edited.]
Dividing the line segment $y$ into a known ratio is not that difficult to construct.