Vector P0C* vector P0B= 1/4[ (vector p0b+p0c) 2-(p0b-p0c) 2]

? = 1/4[(2P0D)^2-(2BD)^2]

? =P0D^2-BD^2

Similarly, vector PC* vector Pb = PD 2-BD 2.

Because vector PC* vector PB, vector P0C* vector P0B.

Namely. PD^2-BD^2》P0D^2-BD^2

That is, PD》P0D.

And because PD reaches a minimum when perpendicular to AB.

That is, P0D is perpendicular to AB.

And because △P0DB is similar to △ABC.

? There is AB/DB=2DB/P0B.

? DB= root number 3

In △PoDB, DP0 2 = (root number 3) 2-12.

? Solution, DP0= root number 2.

h/DP0=CB/DB。

The answer is h=2, root number 2,

That is, the height of the triangle is 2 and the number is 2.