Today, I have compiled 20 classic geometry questions for you, all of which are high-frequency test sites for the senior high school entrance examination. Please share it with your children as soon as possible ~
Classic problem (1)
1, as shown in the figure, O is the center of a semicircle, C and E are two points on the circle, CD⊥AB, EF⊥AB, EG ⊥ Co., Ltd.
Proof: CD = gf.
2. As shown in the figure, P is the interior point of the square ABCD, ∠ Pad = ∠ PDA = 15 degrees.
It is proved that △PBC is a regular triangle.
3. As shown in the figure, it is known that quadrilateral ABCD, A1B1C1D1are all squares, and A2, B2, C2 and D2 are AA 1, BB 1 and CC 1 respectively.
It is proved that quadrilateral A2B2C2D2 is a square.
4. As shown in the figure, in the quadrilateral ABCD, AD = BC, M and N are the midpoint of AB and CD respectively, and the extension line of AD = BC intersects MN at E and F..
Proof: ∠ den = ∠ f.
Classical problem (2)
1. It is known that in △ABC, H is the vertical center (the intersection of height lines on each side), O is the outer center, and OM⊥BC is in m 。
(1) verification: ah = 2om.
(2) If ∠ BAC = 600, verify: ah = ao.
2. Let MN be a straight line outside circle O, let O be OA⊥MN in A, two straight lines drawn from A intersect at B, C, D and E, and straight lines EB and CD intersect at P and Q respectively.
Evidence: Associated Press = AQ.
3. If the above problem translates the straight line MN from outside the circle to inside the circle, the following proposition can be obtained:
Let MN be the chord of circle O, let the midpoint A of MN be two chords BC and DE, and let CD and EB intersect with MN at p and q respectively.
Evidence: Associated Press = AQ.
4. As shown in the figure, take AC and BC of △ABC as one side, make square ACDE and square CBFG on the outside of △ABC, and point P is the midpoint of EF.
Prove that the distance from point P to AB is equal to half of AB.
Classical problem (3)
1, as shown in the figure, quadrilateral ABCD is a square, DE∑AC, AE = AC, AE and CD intersect at F.
Proof: ce = cf
2. As shown in the figure, the quadrilateral ABCD is a square with DE∑AC and CE = CA, and the straight line EC intersects with DA and extends at f 。
Proof: AE = af.
3. Let P be any point on the side of the square ABCD on BC, and PF⊥AP and CF share ∠ DCE equally.
Proof: pa = pf.
4. As shown in the figure, the tangent o of PC is at C, AC is the diameter of the circle, PEF is the secant of the circle, AE and AF intersect with the straight line PO at B and D, which proves that AB = DC and BC = AD.
Classic question (4)
1. It is known that △ABC is a regular triangle, P is a point inside the triangle, Pa = 3, Pb = 4, and PC = 5.
Find the degree of ∠APB.
2. Let p be a point in the parallelogram ABCD and ∠ PBA = ∠ PDA.
Proof: ∠ PAB = ∠ PCB
3. Let ABCD be a convex quadrilateral inscribed in a circle, and prove that AB CD+AD BC = AC BD.
4. In the parallelogram ABCD, let E and F be points on BC and AB respectively, AE and CF intersect at P, and
Ae = cf verification: ∠ DPA = ∠ DPC.
Classic question (5)
1. Let p be any point in positive △ABC with a side length of 1, and L = PA+Pb+PC. Prove:
2. It is known that P is a point in a square ABCD with a side length of 1, and the minimum value of PA+Pb+PC is found.
3.p is a point in the square ABCD, PA = A, Pb = 2A, PC = 3A. Find the side length of a square.
4. As shown in the figure, in △ABC, ∠ ABC = ∠ ACB = 80 degrees, D and E are points on AB and AC respectively, ∠ DCA = 30 degrees, ∠ EBA = 20 degrees, and find the degree of ∠BED.
Answer a question.
Classic problem (1)
4. Connect AC as shown in the figure below, and take the middle point Q to connect QN and QM, then we can get ∠QMF=∠F, ∠QNM =∠ Deng and ∠QMN=∠QNM, and then get ∠ Deng =∠F.
Classical problem (2)
1.( 1) extends AD to even BF, do OG⊥AF,
∠F=∠ACB=∠BHD，
You can get BH=BF, so you can get HD=DF.
And AH = GF HG = GH HD DF HG = 2(GH HD)= 20m.
(2) connecting OB and OC to get ∠BOC= 1200,
So you can get ∠BOM=600.
So OB = 2OM = AH = AO,
Get a license.
Classical problem (3)
Classic question (4)
2. Make a straight line parallel to AD with point P, and choose point E as AE∨DC, BE∨PC.
You can get ∠ABP=∠ADP=∠AEP, and you can get:
AEBP*** circle (one side faces two equal angles).
Get BAP = BEP = BCP, get the certificate.
Classic question (5)
2. Rotate △BPC 60 degrees clockwise to get △PBE of equilateral triangle.
Given that PA PB PC=AP PE EF should be minimized as long as AP, PE and EF are in a straight line,
That is, the following figure: the minimum available PA PB PC=AF.