Test analysis: This question is a geometric proof question, which mainly examines the properties of squares, the properties and judgments of congruent triangles, Pythagorean theorem and other knowledge points. The test question is not difficult, but we should pay attention to the careful calculation in question (3) to avoid mistakes.
To verify DP = DQ, you only need to prove △ ADP △ CDQ to get DP = dq. The key to solve the problem is to find out the condition that the two complementary angles of ∠PDC are equal, that is ∠ ADP = ∠ CDQ, and the two triangles are congruent.
Pe = QE。 We only need to prove △ PDE △ QDE, and we can get the conclusion that DP = DQ plus the bisector with DE ∠ DP=DQ from the conclusion of (1), which can be easily proved by SAS.
(3) From AB: AP = 3: 4 and AB = 6, we can get AP = 8 and BP = 2；; Directly from the conclusions of (1) and (2), let CE = X and PE=8-x, and then use pythagorean theorem to find the edge PE of Rt△PEB, from which the length of EQ can be found, so it is easy to find the area of △DEP.
Test analysis:
(1) proves that ∵ quadrilateral ABCD is a square.
∴DA=DC,∠DAP=∠DCQ=90
∫∠PDQ = 90
∴∠ADP+∠PDC=90
CDQ+∠PDC = 90°
∠ADP=∠CDQ
At △ADP and △CDQ,
∴△ADP≌△CDQ(ASA)
∴DP=DQ
(2) Solution: PE = QE. The evidence is as follows:
∫DE is the bisector of ∞∠PDQ.
∴∠PDE=∠QDE
At △PDE and △QDE,
∴△PDE≌△QDE(SAS)
∴PE=QE
(3) Solution: ∵ AB: AP = 3: 4, AB = 6.
∴AP=8,BP=2，
According to (1), △ ADP △ CDQ is AP = CQ = 8.
It can be known from (2) that △ PDE △ QDE, PE = QE.
Let CE = X, then PE = QE = CQ-ce = 8-X.
In Rt△PEB, BP = 2, be = 6+x and PE = 8-x.
From Pythagorean Theorem: 22+(6+x) 2 = (8-x) 2.
Solution: x =
∴
The area of ∴△DEP is: