In Rt△ABC, AC=6, BC = 8；; Then ab = 10.
∵P is the moving point of AB, which can coincide with A and B (BP of A is 0, BP of B is 10).
But x cannot be equal to 5.
∵ When x=5, P is the midpoint of AB, PM‖AC, and PD‖BC, PD and BC do not intersect, which contradicts the known topic, so the value range of X is 0≤x≤ 10, and x≠5.
It is easy to know △BPH∽△BAC and get:
$\frac{PH}{AC}=\frac{BP}{AB}$，PH=$\frac{AC？ BP } { AB } $ = $ \ frac { 3 } { 5 } $ x；
∴ y = $ \ frac {1} {2} $× 4× $ \ frac {3} {5} $ x = $ \ frac {6} {5} $ x (0 ≤ x≤10 and x ?
(2) when d is on BC,
( 1) ∠ PMB = ∠ B，BP=PM，MH = BH = 2；
At this time △MPD∽△BCA, the result is: $ \ frac {x} {10} = \ frac {2} {8} $,and the solution is $ x = \ frac {5} {2} $.
② When ∠ PMB = ∠ A, △DPM∽△ACB, we get: DP? BA=DM？ BC;
∴ 10x=4×8，x = $ \ frac { 16 } { 5 } $；
When d is on the BC extension line,
Because ∠ PMD > ∠ B, we only discuss ∠ PDM = ∠ B;
When p and a coincide, in Rt△MPD, AC⊥MD, then ∠MAC=∠PDM,
∫tan∠MAC = $ \ frac { 2 } { 3 } $，tanB=$\frac{3}{4}$，tan∠MAC∠PDM；
Therefore, △PDM and △ACB cannot be similar;