(1) ∵ BP ∕ AP =1BD ∕ CD =1∴△ BDE = △ CDE = s1△ BPE = △ ape ∴ s =

S 1∕S2= 1

(2)∵bd∕cd= 1 bp∕ap=n ∴△bde=△cde=s 1 △bpe∕△ape=n

∴s=ns2+s2 △bce∕s=n ∴2s 1∕s=n ∴s=2s 1∕n ∴ns2+s2=2s 1∕n ∴s 1∕s2=n(n+ 1)∕2

(3) let BP ∕ AP = x ∴ 2s1∕ s = x ∴ s = 2s1∕ x: s △ ape = 2 ∴ s △ bpe = 2x ∴.

∴2s 1∕x=2+2x ∴s 1=x( 1+x)∫s△ABC = 24 = s△AEC+s△ape+s△bpe+s△bec

= s+2+2x+2s 1 =(2+2x)+2+2x+2 * x( 1+x)= 24 ∴2x*x+6x+4=24 ∴x*x+3x- 10=0∴(x+5)*(x-2)= 0

∴x=2 or x=-5 (truncated) ∴BP∕AP=2.

I didn't transplant the drawing of the geometric sketchpad to you, so you don't have to! I hope it helps you.