In quadrilateral ABCD, connect AC so that angle ABE= angle ACD and angle BAE= angle CAD.

Triangle ABE is similar to triangle ACD.

So BE/CD=AB/AC, that is, BE*AC=AB*CD (1).

There is also a proportional formula AB/AC=AE/AD.

Angle BAC= Angle DAE.

So triangle ABC is similar to triangle AED.

BC/ED=AC/AD means ED*AC=BC*AD (2).

The same is true of (1)+(2)

AC(BE+ED)=AB*CD+AD*BC

And because BE+ED & gt；; =BD

So this proposition is proved.

reason

Any convex quadrilateral ABCD must have AC BD ≤ ABCD+AD BC. If and only if ABCD has four * * * circles, take the equal sign.

The inverse theorem of Ptolemy's theorem also holds: the sum of the products of two opposite sides of a convex quadrilateral is equal to the product of two diagonal lines, then this convex quadrilateral is inscribed in a circle.

range

Ptolemy inequality: the product of any two opposite sides of a quadrilateral is not less than the product of the other opposite side, if and only if it is a * * * circle or a * * * line.

Simple proof: complex identity: (a-b)(c-d)+(a-d)(b-c)=(a-c)(b-d), take the modulus from both sides, get the inequality, and analyze the conditions for the establishment of the equal sign.

Four points are not limited to the same plane.

On the AD of a line segment, mark B and C in turn, and then AD*BC+AB*CD=AC*BD.

The necessary and sufficient condition for drawing a perpendicular from a point to three sides of a triangle is that the point falls on the circumscribed circle of the triangle.

Prove:

△ABC has a little p on the circumscribed circle, PE⊥AC is in E, PF⊥AB is in F, PD⊥BC is in D, which are connected with DE and DF respectively.

It is easy to prove that P, B, F, D, P, D, C, E and A, B, P and C are * * cycles respectively, so ∠FDP=∠ACP ①, ∫ is ∠ABP's complement) and ∠PDE =∞.

② and ∠ ACP+∠ PCE = 180.

③ ∴∠FDP+∠PDE= 180

(4) that is, the lines of F, D and E***. On the other hand, when the lines of F, D and E***, the circles of A, B, P and E*** can be seen from ④→→→→→→→→→→→ ③→→ ①.