The sum of diagonals of a quadrilateral inscribed in a circle is 180, and any external angle is equal to its internal angle.
As shown in the picture: a picture of a four-point circle.
Figure A: A picture of a four-point * * * circle
The quadrilateral ABCD is inscribed in the circle O, extending the intersection of AB and DC to E, the crossing point E is the tangent EF of the circle O, and AC and BD intersect at P, so there are:
(1)∠A+∠C=π, ∠B+∠D=π (that is, ∠ DAB+∠ DCB = π in the figure, ∠ ABC+∠ ADC = π).
(2)∠DBC=∠DAC (the circumferential angles of the same arc are equal).
(3)∠ADE=∠CBE (the outer angle is equal to the inner diagonal, which can be obtained from (1) and (2))
(4)△ABP∽△DCP (three internal angles of two triangles are equal, which can be obtained from (2))
(5)AP*CP=BP*DP (chord theorem)
(6)EB*EA=EC*ED (Secant Theorem)
(7) ef? = EB*EA=EC*ED (Secant Theorem)
(8)AB*CD+AD*CB=AC*BD (Ptolemy theorem)
Decision theorem
Method 1: connect the four points of the proved * * * circle into two triangles with the base of * * *, and both triangles are on the same side of the base. If it can be proved that their vertex angles are equal, then these four points of the circle are certain.
(It can be said that if the angle between two points on the same side of the line segment and two points at both ends of the line segment is equal, then these two points and four points at both ends of the line segment are * * * circles. )
Method 2: Connect the four points of the proved circle into a quadrilateral. If it can be proved that its diagonal is redundant or its outer angle is equal to the inner diagonal of its adjacent complementary angle, then a four-point circle can be sure.
(It can be said that if four points on a plane are diagonally complementary or an outer angle is equal to its inner diagonal, then these four points are * * * circles. )
ptolemy's theorem
Ptolemy Theorem: If ABCD has four * * * circles (ABCD is on the same circle in turn), then AB*DC+BC*AD=AC*BD.
Example: It is proved that any positive integer n has n points, so that the distance between all points is an integer.
Answer: induction. We prove a stronger theorem by induction: for any n, there are n points that make the distance between all points an integer, these n points are * * * circles, and the two points are the two ends of a diameter. N= 1, n=2 is easy. When n=3, a pythagorean triangle with an integer side length is enough: for example, triangles with sides of 3, 4 and 5. We find that the longest side of such a three-point circle is a diameter. Let's prove that n+ 1 is true for n ≥ 3. Suppose the diameter is r (integer). Find an integer pythagorean triangle ABC (side length a
Siemsen line
Sim's Theorem: If any point on the circumscribed circle of a triangle is perpendicular to three sides or their extension lines, the line is perpendicular. This line is usually called the siemsen Line.
Decision 1
Choose three points from the four points of a circle to make a circle, and then prove that the other point is also on this circle. If we can prove this, we can determine the four points of the circle.
Inference: The distances from the points of the proved * * * circles to a certain point are all equal, so they are judged to be * * * circles. That is, the perpendicular lines of the three sides of the connected quadrilateral have intersections, and these four * * * circles can be determined.
Decision 2
1: Connect the four points of the proved * * * circle into two triangles with the base of * * *, and both triangles are on the same side of the base. If we can prove that their vertex angles are equal (the relative circumferential angles of the same arc are equal), we can be sure of these four points of the circle.
2. Connect the four points of the proved * * * circle into a quadrilateral. If it can be proved that its diagonal is a complementary angle or that one of its outer angles is equal to the inner diagonal of its adjacent complementary angle, the four-point * * * circle can be affirmed.
See the proof method below.
Decision 3
Connect the four points of the proved * * * circle into two intersecting line segments. If it can be proved that the products obtained by dividing these two line segments by their intersection points are equal, the four-point * * * circle (the inverse theorem of the intersecting chord theorem) can be affirmed; Or connect the four points of the proved * * * circle in pairs and extend the two intersecting line segments. If we can prove that the product of two line segments from the intersection to the two endpoints of one line segment is equal to the product of two line segments from the intersection to the two endpoints of another line segment, we can be sure that these four points are also * * * circles. (Inverse theorem of secant theorem)
The above two theorems are collectively called the inverse theorem of circular power theorem, that is, four points of ABCD connect AB and CD respectively, and the intersection of them (or their extension lines) is P. If PA*PB=PC*PD, ABCD has four points * * * circle.
Proof: connect AC, BD, PA * PB = PC * PD.
∴PA/PC=PD/PB
* armoured personnel carriers = BPD
∴△APC∽△DPB
When P is on AB and CD, it is similar to ∠A=∠D, and A and D are on the same side of BC. According to Method 2, we can know that ABCD is a four-point circle.
When P is on the extension line of AB and CD, we can get ∠PAC=∠PDB from the similarity, and A and D are on the same side of BC. Similarly, according to method 2, we can know the four-point * * * circle of ABCD.
Decision 4
In quadrilateral ABCD, if AB*CD+AD*BC=AC*BD, that is, the sum of the products of two opposite sides is equal to the product of diagonal lines, then ABCD is a four-point * * circle. This method can be obtained from the inverse theorem of Ptolemy theorem.
Inverse theorem of Ptolemy's theorem: For any convex quadrilateral ABCD, there is always AB*CD+AD*BC≥AC*BD, and the equal sign is established if ABCD has four * * * circles.
As shown in the figure, do △APB∽△DCB in quadrilateral (just do △ PAB = △ CDB, △ PBA = △ CBD).
By similar ∠ABP=∠DBC, ∠BAP=∠BDC.
∴∠ABP+∠PBD=∠DBC+∠PBD
That is ∠ Abd =∠PBC.
And similar AB:BD=PB:CB=AP:CD.
∴AB*CD=BD*AP,△ABD∽△PBC
∴AD:BD=PC:BC, that is, AD*BC=BD*PC.
When the two equations are added, AB*CD+AD*BC=BD*(PA+PC)≥BD*AC, and the necessary and sufficient condition for the equal sign is APC three-point * * * line.
The APC*** line indicates ∠BAP=∠BAC, and ∠BAP=∠BDC, ∴∠BAC=∠BDC.
According to the decision 2- 1, ABCD is a four-point circle.
Decision 5
Inverse theorem of siemsen's theorem: If a point is on the projective straight line of three sides of a triangle, it is on the circumscribed circle of the triangle.
There is a △ABC, where P is a point on the plane different from ABC, and the vertical lines passing through P are L, M and N respectively. If the straight line of l, m and n is * * *, then p is on the circumscribed circle of △ABC.
As shown in the figure, PM⊥AC,PN⊥AB,PL⊥BC and L, N and M are on the same straight line.
Connect PB, PC, ∫≈pl b+∠PNB = 90+90 = 180.
∴PLBN quadrangle * * * circle
∴∠PLN=∠PBN, that is ∠PLM=∠PBA.
Similarly, ∠PLM=∠PCM, that is ∠PLM=∠PCA=∠PBA.
According to the decision 2- 1, p is on the circumscribed circle of △ABC.
I hope it can help you solve the problem.