(1) First prove that △ABC is a right triangle with ∠ACB as the right angle;

(2) Make symmetrical figures and calculate the area of △ABC.

Detailed answer:

(1) Intercept AD=AC on line AB

AB = 2AC，AD=AC

∴ DB=DC

∴ ∠ABC=∠DCB

AD = AC, and the angle between ad and AC ∠ BAC = 60.

△ ADC is an equilateral triangle.

∴ ∠ADC = 60

∫∠ADC is the external angle of isosceles △DBC.

∴∠ADC = ∠ABC + ∠DCB

= 2∠ABC

60 = 2 ∠ ABC

∴ ∠ABC = 30

And < BAC = 60.

∴, ∠ ACB = 90 in △ABC.

(2) Make the symmetry point of point P about AC P 1,

Make the symmetric point P2 of point P about AB,

Make P3, the symmetry point of P about BC,

Even AP 1, AP2 and P 1P2 are known from symmetry:

△AP 1P2 is an isosceles triangle, AP 1 = AP2 = √3, P 1P2 = 3, ∠ p 1ap2 = 2 ∠ BAC = 120.

It is easy to find the area of isosceles △P 1AP2: S△P 1AP2 = (3√3)/ 4.

Similarly, △BP2P3 is an equilateral triangle, BP2 = BP3 = P2P3 = 5, ∠ p2bp3 = 2 ∠ ABC = 60.

It is easy to find the area of equilateral △BP2P3: S△BP2P3 = (25√3)/ 4.

Even CP 1 even CP3,

Then CP 1 = CP = 2, CP3 = CP = 2, ∠ P 1cp3 = 2 ∠ ACB = 180, that is, lines P 1, C and P3 * *.

In △P 1P2P3，P 1P2 = 3，P 1P3 = 4，P2P3 = 5。

∴ △P 1P2P3 is Rt△, and its area is: S Rt△P 1P2P3 = 6.

∴s△p 1ap2+s△bp2p 3+s rt△p 1p2p 3

=(3√3)/ 4 +(25√3)/ 4 + 6

= 7√3 + 6

According to symmetry:

S△AP2B = S△APB

S△BP3C = S△BPC

S△AP 1C = S△APC

∴ S△APB + S△BPC + S△APC

=( 1/2)×(S△p 1ap 2+S△bp2p 3+S Rt△p 1p2p 3)

= ( 1/2)× (7√3 + 6)

= (7√3 + 6)/ 2

That is, the area of △ABC is (7√3+6)/ 2.