Prove the condition of similar triangle, AAA.

You can use the sin law to prove that.
Sin Law tell us that in the triangle ABC,
AB/sin C = BC/sin A = AC/sin C

Let there is two triangle PQR and XYZ, and angle P=X, Q=Y and R=Z.

In the triangle PQR,we have:
PQ/sin R = QR/sin P = PR/sin Q..............(1)

In the triangle XYZ, we have:
XY/sin Z =YZ/sin X = XZ/sin Y....................(2)

(1)/(2),we have:
PQsinZ/XYsinR = QRsin X/YZsinP = PRsin Y/XZsin Q...........(3)

since angle P=X, Q=Y and R=Z,
sin P=sin X, sin Q=sin Y and sin R=sin Z..................(4)

Put (4) into (3), we have:
PQ/XY = QR/YZ = PR/XZ

All the sides are in proportion, so triangle PQR is similar to triangle XYZ.

http://i207.photobucket.com/albums/bb173/kkkpopup/solution.gif

This method is suitable to Form 3 students who are not expected to know Sine Law.

Yes, "3 sides proportional as well as equiangular" (both are true) is the definition of similar triangles.