In the triangle ABC, if the angle ACB is a right angle, BC = 3, AC = 4, and P is a point on AB, then the product of the distances from P to AC, AB is the maximum___________

In the triangle ABC, if the angle ACB is a right angle, BC = 3, AC = 4, and P is a point on AB, then the product of the distances from P to AC, AB is the maximum___________

Let AP = X
The distance from P to AC is 3x / 5
The distance between P and ab is 4 (5-x) / 5
Then the product of the distance from point AB to point AC-5 / P = 25

In the triangle ABC, C = 90 °, AC = 4, BC = 3, and point P is on edge AB, try to find the maximum product of the distances from point P to edge AC, BC
1. In the triangle ABC, where ∠ C = 90 °, AC = 4, BC = 3, point P is on edge AB, try to find the maximum product of the distances from point P to edge AC, BC
2. The distance from the moving point P to the vertex A is twice of that to the fixed point B, and | ab | = 2A (a > 0)
If you think hard but fail, ask for advice

The first problem is to take C as the origin and AC BC as the coordinate axis to build a coordinate system
The second problem takes the midpoint of AB as the origin and ab as the x-axis to build a rectangular coordinate system

Given that in the triangle ABC, the angle ACB = 90 degrees, BC = 1, AC = 2, P is a moving point on AB, then the maximum value of the distance product from point P to AC, BC is?

Connection CP.PD ⊥ AC in D, PE ⊥ BC in E ⊥ s △ PBC = 1 / 2BC · PE s △ PAC = 1 / 2Ac · PDS △ ABC = 1 / 2Ac · BC = 1 ⊥ s △ PBC + s △ PAC = s ⊥ ABC ⊥ 1 / 2 · PE + PD = 1 ⊥ PE + 2PD = 2 ⊥ PE > 0 2PD > 0 ⊥ PE + 2PD = 2 ≥ 2 √ (PE · 2PD) ⊥ PD · PE ≤ 1 / 2 ⊥ distance product of point P to AC, BC