AB+BC = 30 AB-BC = 8 AB = 19 BC = 1 1

2, 1)≈BAF =∠EAD Because of the nature of parallel lines, two lines are parallel and have the same angle.

∠EAD=∠F =∠FAB=∠E In a triangle, CEF ∠F =∠E So a triangle is an isosceles triangle.

2)∠EAD=∠F =∠FAB=∠E So the triangle BAF and DAE are isosceles triangles.

FB=AB AD=DE The parallel perimeter of an isosceles triangle is AB+BC+CD+AD, and the two waists are FB+BC+CD+AD.

That is, the length of the two waists is equal to the perimeter of the parallelogram.

3. It is proved that ∠C=∠DFE AD is parallel to BC ∠AFE=∠CEF ∠BEF=∠DFE is folded to point F because of point C.

Because ∠ AFE+∠ DEF = 180, ∠ dfe+∠ cef =180 ∠ FDC+∠ c =180.

And because ∠C=∠DFE, so ∠DFE =∠ C quadrilateral CEFD is a parallelogram with DC parallel to EF.

2)AF is a parallelogram parallel to BE AB and EF. BEF ∠ BEA = ∠ FEA ∠ FED = ∠ CED.

∠BEA+∠FEA+∠FED+∠CED = 180∠DEA =∠AEF+∠DEF = 90

4、s = 2 *( 1/2)* BD * ad = 3 * 4 = 12 ac=2oa=2√(3^2+4^2)=2√ 13

5. Seek its value.

6. ∠ ACB = 90 ∠ D "CE" = 60 ∠ E "CB =15, so ∠BCO = 45 =∠ACO =∠ Cao ∠BCO.

So ∠ AOC = 90.

∠FOD " = 90∠OD " F = 30 so∠AFD " = 60 so∠ofe " = 180-∠AFD " = 120。

In the isosceles right triangle AOC, the hypotenuse AC=3√2, so AO=CO=3 and CD"=7 OD"=4.

In the right triangle AOD ",the hypotenuse AD" = 5.

7. Because ∠ B =120 ∠ A = 60 ∠ AED = 90 ∠ ADE = 30.

Similarly ∠ CDF = 30 ∠ ADC = ∠ B =120 ∠ def =120-∠ ade-∠ CDF = 60.