Euler's Theorem In Geometry

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In geometry, **Euler's theorem**, named after Leonhard Euler, states that the distance *d* between the circumcentre and incentre of a triangle can be expressed as

where*R* and *r* denote the circumradius and inradius respectively (the radii of the above two circles).

From the theorem follows the**Euler inequality**:

## Proof

Let *O* be the circumcentre of triangle *ABC*, and *I* be its incentre, the extension of *AI* intersects the circumcircle at *L*, then *L* is the mid-point of arc *BC*. Join *LO* and extend it so that it intersects the circumcircle at *M*. From *I* construct a perpendicular to AB, and let D be its foot, then *ID* = *r*. It is not difficult to prove that triangle *ADI* is similar to triangle *MBL*, so *ID* / *BL* = *AI* / *ML*, i.e. *ID* × *ML* = *AI* × *BL*. Therefore 2*Rr* = *AI* × *BL*. Join *BI*, because

therefore angle*BIL* = angle *IBL*, so *BL* = *IL*, and *AI* × *IL* = 2*Rr*. Extend *OI* so that it intersects the circumcircle at *P* and *Q*, then *PI* × *QI* = *AI* × *IL* = 2*Rr*, so (*R* + *d*)(*R* − *d*) = 2*Rr*, i.e. *d*<sup>2</sup> = *R*(*R* − 2*r*).

## External links

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- <math> d^2=R (R-2r) ,</math>

where

From the theorem follows the

- <math>R ge 2r. </math>

- angle
*BIL*= angle*A*/ 2 + angle*ABC*/ 2,

- angle
*IBL*= angle*ABC*/ 2 + angle*CBL*= angle*ABC*/ 2 + angle*A*/ 2,

therefore angle

<math></math>

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