2. Using the area invariance, the area of this triangle is calculated in two ways. If we know the relationship between the two bottoms, then the relationship between the two highs will be known.
3. Because ∠A is equal to 30 and ∠C is equal to 60, according to the theorem of triangle interior angle sum ∠BEC is equal to 80.
4: It becomes a quadrilateral, and the sum of the internal angles of the quadrilateral is 360 degrees.
5: Because the sum of the internal angles of the quadrilateral is 360, ∠B+∠D is equal to180, so ∠ADF+∠FBE is equal to 90 (because these two angles are half of the corresponding large angle, the sum is half), and ∠ ADF.
Do you understand these questions? If you don't understand, please say it first. I want to do the following questions.
6. The formula of the sum of polygon internal angles is (n-2) × 180, where n is the number of sides, making it equal to 2220, and the result is n= 14.3, plus an angle, that is, 15 deformation.
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7;
By making these three auxiliary lines, we can get C+D+E = 360, while A+B is equal to 180, so it is a quadratic relationship.
8; 2000 is greater than 13 and less than 14, so it is 13 deformation.
9: Connect CF, because AF//CD, angle AFC= angle FCD. In quadrilateral ABCF, angle A plus B plus AFC plus FCB=360, so AFC plus FCB=360- 100- 140, we get: FCD plus FCB= 120.
Same reason? Connecting AD, you can get the angle D= 100.
10: connect BC.
BE is the bisector of ∠ABD and CF is the bisector of ∠ACD.
∴∠ABE=∠DBE=
1
2
∠ABD，∠ACF=∠DCF=
1
2
∠ACD，
∠ BDC = 140，∠ BGC = 1 10，
∴∠DBC+∠DCB=40，∠GBC+∠GCB=70，
∴∠EBD+∠FCD=70 -40 =30，
∴∠ABE+∠ACF=30，
∴∠ Abe+∠ ACF+∠ GBC+∠ GCB = 70+30 =100, that is ∠ ABC+∠ ACB = 100.
∴∠A=80。
It's finally finished, you know?