1. As shown in the figure, do you want to make it? ABCD becomes a rectangle, and the condition to be added is ()
A.ab = BC b .∠ABC = 90°c .∠ 1 =∠2 d.ac⊥bd
2. As shown in the figure, in △ABC, AD⊥BC at point D, de∨AC at point E and df∨ab at point F connect DE and FD. When △ABC satisfies the condition, the quadrilateral AEDF is a rectangle.
3. As shown in the figure, in? In ABCD, point m is the midpoint of CD edge, and am = BM. Prove that the quadrilateral ABCD is a rectangle.
4. In math activity class, teachers and students judge whether a quadrilateral door frame is rectangular. The following is a plan drawn up by four students in a cooperative learning group, of which the correct one is ().
A. measure whether the diagonal lines are equally divided. B. measure whether the two groups of opposite edges are equal respectively.
C, measuring whether a set of diagonal angles are right angles d, and measuring whether three corners of a quadrilateral are right angles.
5. The quadrilateral surrounded by the bisector of the inner corner of the parallelogram is ()
A. arbitrary quadrilateral B. parallelogram C. rectangle D. none of the above is right
6. As shown in the figure, in △ABC, AB = AC, AD and AE are bisectors of ∠BAC and ∠BAC's external angles BE⊥AE, and the vertical foot is E. 。
(1) verification: large ⊥ AE;
(2) Try to judge whether AB and DE are equal? And prove your conclusion.
7. The diagonal AC and BD of quadrilateral ABCD are equally divided. To make it a rectangle, the condition you need to add is ().
A.AB=CD B.AC=BD C.AB=BC D.AC⊥BD
8. As shown in the figure, AB = AC, AD = AE, DE = BC, ∠ Bad = ∠ CAE. It is proved that the quadrilateral BCDE is rectangular.
9. As shown in the figure, the quadrilateral ABCD is a parallelogram, extending AD to point E, making DE = AD, connecting EB, EC and DB, and adding a condition, it is () that cannot make the quadrilateral DBCE rectangular.
A.AB=BE B.BE⊥DC C.∠ADB=90 D.CE⊥DE
10. In quadrilateral, ① ab = cd; , diagonal AC and BD intersect at point O, from ① AB = CD; ②AB∑CD； ③OA = OC； ④OB = OD； ⑤AC = BD； ⑥ ∠ ABC = 90. Of these six conditions, three can be selected to deduce that quadrilateral ABCD is rectangular, such as 12⑤→ quadrilateral ABCD is rectangular. Please write two more combinations that meet the requirements:.
1 1. As shown in the figure, in the rectangular ABCD, m is the midpoint of the AD side, and p is the upper point of pf⊥mb pe⊥mc BC. When AB and BC meet the conditions, the quadrilateral PEMF is a rectangle.
12. As shown in the figure, in the parallelogram ABCD, points E, F, G and H are on the sides of AB, BC, CD and AD respectively, AE = CG and Ah = CF 。
(1) Verification: the quadrilateral EFGH is a parallelogram;
(2) If AB = AD and AH = AE, it is proved that the quadrilateral EFGH is a rectangle.
13. As shown in the figure, in Rt△ABC, ∠ C = 90, AC = 8°, BC = 6°, and point P is any point on AB. Let PD⊥AC be at point D, PE⊥CB be at point E and connect DE, then the minimum value of DE is _ _.
14. As shown in the figure, in △ABC, point O is a moving point on the side of AC, and the intersection point O is a straight line MN∨BC. Let the bisector of MN intersecting with ∠ACB be at point E, and the bisector of external angle intersecting with ∠ACB be at point F. 。
(1) verification: OE = of
(2) If CE = 12 and CF = 5, find the length of OC;
(3) When the point O moves to where on the AC side, is the quadrilateral AECF a rectangle? And explain why.
Reference answer
1.B
2.∠BAC=90
3. Yi Zheng △ AMD △ BMC (SSS), ∴∠ C = ∠ D. And ∠ C+∠ D = 180,
∴∠c =∠d = 90°, ∴ parallelogram ABCD is a rectangle.
4.D
5.C
6.( 1)∵AD shares ∠BAC, ∴ Bad = 12 ∠ BAC, while ∵AE shares ∠BAF.
∴∠ BAE = 12 ∠ BAF，∠∠∠BAC+∠BAF = 180，∴∠ Bad+∠ BAE = 65438+。 So ∠ ADB = 90, ∵BE⊥AE, ∴∠ AEB = 90, ∠ DAE = 90, so the quadrilateral AEBD is a rectangle. Company.
7.B
8. Links BD, EC, ∠∠bad =∠CAE, ∴∠ bad-∠ BAC = ∠ CAE-∠ BAC, ∴∠ BAE = ∠ CAD, and ∞
9.B
10.①②⑥ ③④⑥
1 1.AB= 12BC
12.( 1) In the parallelogram ABCD, ∠ A = ∠A=∠C, ∠ B = ∠ D, and ∵ AE = CG,
Ah = CF, ∴△AEH≌△CGF(SAS), ∴ EFGH = GF, in the parallelogram AB=CD, AB = CD, AD = BC, ∴ AB-AE = CD-CG, AD-AH = BC-CF, that is, be.
(2) AB=CD, AB∑CD and AB = CD in the parallelogram.
Let ∠ A = α, then ∠ D = 180-α, AE = Ah,
∴∠AHE=∠AEH= 180 -α2=90 -α2，ad = ab = cd，
Ah = AE = CG, ∴ ad-ah = CD-CG, namely DH = DG,
∴∠dhg=∠dgh= 180-( 180-α)2 =α2，
∴∠EHG= 180 -∠DHG-∠AHE=90，
The quadrilateral EFGH is a parallelogram, and the quadrilateral EFGH is a rectangle.
13.4.8
14.( 1)∫cf shares ∠ACD, while Mn∨BD, ∴∠ ACF = ∠ FCD = ∠ CFO, ∴ of = oc. It can also be proved that: oc.
(2) From (1), we can know that: of = oc = OE, ∴∠ OCF = ∠ OFC, ∠ OCE = ∠ OEC, ∴∠ocf+∠oce =∞.
(3) When the point o moves to the AC midpoint, the quadrilateral AECF is rectangular. The reason is that AE and AF are connected, and OE = of is known from (1). When the O point moves to the AC midpoint, there are OA = OC, quadrilateral AECF is parallelogram, ECF is ≈ 90, ∴.