Current location - Training Enrollment Network - Mathematics courses - A geometric mathematics problem about right triangle
A geometric mathematics problem about right triangle
(1) connects AO, BO and co, which is known from Pythagorean theorem.

The left side of the equation is equal to

OA? -What? +OB? -An overdose? +OC? -OE?

The right side of the equation is equal to

OB? -What? +OC? -An overdose? +OA? -OE?

The two formulas are equal, so the conclusion holds.

⑵ Using the conclusion of the previous question, all the formulas on the right side of the equation are moved and obtained by the square difference formula.

(AF+FB)(AF-FB)+(BD+DC)(BD-DC)+(CE+EA)(CE-EA)= 0

Because af+FB = BD+DC = ce+ea > 0 (because the equilateral tangents of three sides of a regular triangle are greater than 0

So there is

AF-FB+BD-DC+CE-EA=0

Finishing can get the required formula.