We know that the line AB = 8, and there is a point P on the plane. (1) if AP = 5, what is Pb equal to, P is on the line AB?

Draw a line and mark it with P,
Because AB = 8, AP = 5
So when Pb = ab-ap = 8-5 = 3,
P is on line ab

If AB = 8, there is a point P on the plane. (1) if AP equals 8, P is on AB (2). If AB = 5, what does Pb satisfy
Given the line AB = 8, there is a point P on the plane
(1) If AP equals 8, point P is on line ab
(2) If AB = 5 and Pb satisfies what conditions, point P is not on line ab
(3) When PA = Pb, the position of point P is determined, and the size of PA + Pb and ab is compared

When AP is 8 and crosses a point on AB, draw a circle with 8 as radius, ∵ AB = 8 ∵ intersect AP with B. the second problem is AP = 5, draw a circle with a as center and 5 as radius, because P is not on line AB, and because the sum of two sides of the triangle is greater than the third side, so AP + BP > AB, so Pb > 3 (3) when PA = PBP is not on line AB, △ ABP is a

A. B is two points on the plane, ab = 10cm, P is one point on the plane, if PA + Pb = 20cm, then p ()
A. It can only be outside the line ab. it can only be on the line ab. C. It can't be on the line ab. D. It can't be on the line ab

As shown in the figure, according to the figure, point P can be on line AB or outside line AB, but not on line ab

It is known that α‖ β. Point P is a point outside the plane α and β. Three non coplanar rays PA, Pb and PC are introduced from point P, which intersect plane α at point a B C and plane β at a 'B' C 'respectively to find △ ABC ∽ a' B 'C'

From the meaning of the title, we can see that P, a, B, a 'and B' are in the same plane, because α‖ β knows ab ‖ a 'B', so PA / PA '= AB / a' B '
Similarly, it can be proved that PA / PA '= AC / a'c', so AB / a'B '= AC / a'c'
Similarly, it can be proved that AB / a'B '= AC / a'c' = BC / b'c '
So △ ABC ∽ a'b'c

We know that the line AB = 8, and there is a point P on the plane. (1) if AP = 5, what is Pb equal to, P is on the line AB?