①-②, z=80 ④, substitute ③, y=20, substitute ②, x=80.

2. Let the three sides of an isosceles triangle be X, X and Y, then x+y= 12①, x-y=6②, ①+②, 2x= 18, x=9, ①-②2y=6, y=3, then the circumference of the triangle is

3. Connect the CD, and extend the intersection point AB to point E. In the delta △ADC, the external angle ∠ADE=∠DAC+∠ACD①, in the same way ∠BDE=∠DBC+∠BCD②, and ∠ ADB = ∠ ADE+.

4. If the quadrilateral inner angle is 360, then ∠ EOF+∠ A+∠ AEO+∠ AFO = 360, while AE and CF are higher, so ∠ AEO = ∠ AFO = 90, and ∠.

5.∠BAC=∠ACD+∠D, ∠DCE=∠B+∠D, that is ∠B=∠DCE-∠D,

And CD share ∠ACE, ∠ACD=∠DCE, so ∠ BAC >; ∠ACD & gt； ∠B, that is ∠ BAC > ∠B

6. Let the number of one side be n and the number of the other side be 2n.

The sum of internal angles is (n-2) * 180 and (2n-2) * 180 respectively.

That is, (n-2)/(2n-2)= 1/4, n=3, 2n=6.

Namely triangle and hexagon.