1. First, it is proved that1/pc =1/2 (1/pa+1/pb) can make AC=x, PB=3+x, and substitute PC=2+x to get x=[ root.

2. Now let's prove that1/PC =1/2 (1/pa+1/Pb):

Let PO and ST intersect at D, and let the intersection of OH perpendicular to PB be H.1/PA+1/Pb = (PA+Pb)/(PA * Pb). Obviously, there are PA * Pb = PS 2 (the square of PS) and PA+PB=2*PH, so the original proposition becomes.

1/PC = pH/PS 2 or PC * pH = PS 2 (1).

Note that the triangular PSD is similar to POS, and PS 2 = PD * PO (2) is obtained.

According to (1)(2), as long as it is proved,

PC*PH=PD*PO

Then notice that the triangle PCD is similar to POH, and immediately push the formula, and the proof is over!

You should have a good foundation. Can I see this? Send me a message next time you have any questions. There is nothing I can't solve.