It is known that the area of parallelogram ABCD is s, and the area of triangle PAB and triangle PCD are S1 and S2 respectively If point P is outside the parallelogram ABCD, then S1 + S2 -- half s Please explain why

The sum of their heights is exactly the height of the parallelogram, and the bottom edge is equal to the length of the bottom edge of the parallelogram. However, your P point must fall between the two bottom edges. Otherwise, it will be a negative area

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1. As shown in the figure, in the pyramid s-abcd, the bottom surface ABCD is a parallelogram, the side SBC ⊥ the bottom surface ABCD, ∠ ABC = 45 °, SA = sb, proving that SA ⊥ BC
2. Given any parallelogram ABCD, e is the midpoint of AD, f is the midpoint of BC, the proof is: ab + DC = 2ef
3. In the parallelogram ABCD, m and N are the midpoint of DC and BC respectively. The known vectors am = C and an = D are used to represent the vectors AB and AD
4. In the parallelogram ABCD, passing through point C makes CE ⊥ CD and intersects ad at point E, Next, If ad = 6, tanb = 4 / 3, AE = 1, let CP1 = x, s △ p1f1c1 = y, find y and X, find the functional relationship between X and y, and write out the value range of independent variable x
5. In the parallelogram ABCD (Xining City, 2008), the bisector ce of angle BCD intersects the edge ad at E (Xining City, 2008) 23. As shown in Figure 10, it is known that in the parallelogram ABCD, the bisector ce of angle BCD intersects ad at e, the bisector BG of angle ABC intersects CE at F, and ad at g. verification: AE = DG
6. In the parallelogram ABCD, it is known that the bisector ce of ∠ BCD intersects ad at e, the bisector BG of angle ABC intersects CE at f and ad at g. the proof is AE = dg
7. If the bisectors of parallelogram ABCD, ab = 5, ad = 8, angles C and D intersect ad, BC and points E and f respectively, and AF is perpendicular to BC, the solution of CE is obtained
8. In the parallelogram ABCD, ab ∥ CD, ad ∥ BC, e and F are on AD and CD respectively, and CE = AF. CE and AF intersect at point P, and In the parallelogram ABCD, ab ∥ CD, ad ∥ BC, e and F are on AD and CD respectively, and CE = AF, CE and AF intersect at point P, and Pb bisection ∠ APC is proved
9. In square ABCD, f is the midpoint of AD, BF and AC intersect at point G, then the area ratio of triangle BGC to quadrilateral CGFD is
10. In square ABCD, f is the midpoint of AD, BF and AC intersect point G, then the area ratio of triangle BGC and quadrilateral CGFD is? How to write the correct procedure?
11. If the area of parallelogram ABCD is 100, then s △ PAB + s △ PCD =
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13. The two diagonals of quadrilateral ABCD are perpendicular to each other and intersect at O. given AC = 4cm, BD = 5cm, find the area of quadrilateral ABCD
14. Given the parallelogram ABCD, make the equilateral triangle ade and equilateral triangle BCF outward with AD and BC, connect be and DF, and find be = DF
15. Given the parallelogram ABCD, take ad BC as the edge, make positive △ ade and positive △ BCF outside the parallelogram, connect BD and EF, and they intersect at O, and prove EO = fo, do = Bo There is no picture
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19. As shown in the figure, ABCD is a parallelogram, EF is parallel to ac. if the area of the triangle is 4 square centimeters, calculate the area of the triangle CDF?
20. In the parallelogram ABCD, make a straight line AF through a, cross BC and E, and cross DC extension line to F. it shows that the area of triangle ABF and ade is equal