Analysis of test questions: Question types: According to the available △ ade △ FCE, we can draw the conclusion that S△ADE=S△FCE;
Problem transfer: according to the conclusion of the problem situation, it can be concluded that when the straight line rotates to point P, that is, the midpoint of MN, S△MON is the smallest, and when it passes through point M, it is MG∑OB, and EF satisfies G. It can be concluded from the properties of congruent triangles;
Practical application: As shown in Figure 3, let PP 1 ⊥OB, MM 1 ⊥OB, and the vertical feet are respectively P 1 and M 1, and then draw a conclusion from the trigonometric function value according to the conditions;
Extension and extension: It is discussed that the straight line L passing through point P intersects with a set of opposite sides OC and AB of quadrilateral OABC at points M and N, respectively, and extends the intersection of OC and AB at point D, so that the area of △OAD can be obtained under the condition of AD=6, and then the maximum value can be obtained according to the conclusion of problem migration;
The straight line L passing through point P intersects with another group of opposite sides CB and OA of quadrilateral OABC at m and n, respectively, and the X axis of CB extends to T. From the coordinates of B and C, the analytical expression of straight line BC can be obtained, and then the coordinates of T can be obtained, and then the area of △OCT can be obtained. Then get the maximum value from the conclusion of problem migration, and draw a conclusion through comparison.
Problem situation: ∵AD∨BC,
∴∠DAE=∠F,∠D=∠FCE.
Point e is the midpoint of the DC edge,
∴DE=CE.
∫ In △ Ade and △FCE,
,
∴△ADE≌△FCE(AAS)，
∴S△ADE=S△FCE，
∴S quadrilateral ABCE +S △ADE =S quadrilateral ABCE +S △FCE,
That is, s quadrilateral ABCD = s△ABF；;
Problem transfer: When the straight line rotates to the point where P is the midpoint of MN, S△MON is the smallest, as shown in Figure 2.
Another straight line passing through point P, EF intersects OA and OB at points E and F, let PF < PE, let M intersect MG∨OB, and EF intersects G,
From the problem situation, it can be concluded that when P is the midpoint of MN, the S quadrilateral MoFG = S △ mon.
∫S quadrilateral mofg < s △ eof,
∴S△MON