If R and r is the Circumradius and Inradius of a non-degenerate triangle then due to Euler we have an Inequality stated as and the equality holds when the triangle is equilateral. This ubiquitous inequality occurs in the literature in many different equivalent forms [4] and also Many other different simple approaches for proving this inequality are known. (some of them can be found in [2], [3], [5], [17], and [18] ). In this article we present a proof for this Inequality based on two basic lemmas, one is on the fact “Among all the pedal’s triangles, the pedal triangle whose area is one fourth of area of the reference triangle has maximum area” and other is related to the “area of Ex-Touch triangle”.
Notations:
Let ABC be a triangle. We …show more content…
Ex-Touch triangle: If D, E, F are the points of contact of excircles opposite to the vertices A, B, C with the sides BC, CA, AB respectively then the triangle formed by joining the points D, E, F is called as Extouch triangle. It is well known that Ex-Touch triangle is a pedal’s triangle with respect to Bevan point of [7].
5. Bevan point: Bevan point(V) is the Circumcenter of the excentral triangle. It is named in honor of Benjamin Bevan. The Bevan point is the reflection of Incenter in the Circumcenter and it is also the reflection of orthocenter in the Spieker center. The Bevan point(V) is the midpoint of line segment joining the Nagel point and de-Longchamps point.The Bevan point(V) is a triangle center and it is listed as the point X(40) in Clark Kimberling’s Encyclopedia of Triangle Centers[15].
Lemma – 1 If is the area of pedal’s triangle with respect to the any arbitrary point P, and if is the area of reference triangle then prove that Among all the pedal’s triangles, the pedal triangle with respect to some point P whose area is one fourth of the area of the reference triangle, has maximum area. That is . …show more content…
Hence proved.
Lemma – 2
The Ex-Touch triangle of reference triangle acts as the pedal’s triangle with respect to the Bevan point and the area of Ex-Touch triangle is
Proof:
Let D, E, F are the points of tangency of excircles opposite to the vertices A, B, C of with the sides BC, CA, AB and if IA, IB, IC are the excenters opposite to the vertices A, B, C then IAD, IBE, ICF are perpendicular to the sides BC, CA, AB respectively.
Let V is the point of intersection of extensions of IAD, IBE, ICF respectively, now It is enough to prove that the point V is the circumcenter of Excentral triangle IAIBIC that is V is the Bevan point, it results that Ex-Touch triangle is the pedal’s triangle of with respect to Bevan point[1]. Now by angle chasing and using little algebra we can prove that , and
It implies that IAV = IBV = ICV
So V is the Circumcenter of Excentral triangle IAIBIC. Hence Ex-Touch triangle is the pedal’s triangle of with respect to Bevan point.
Now let us find the area of Ex-Touch triangle,
We are familiar with the result that “In any triangle cevian divides the triangle into two triangles whose ratio between the areas is equal to the ratio between corresponding bases”.
So and it implies
Notations:
Let ABC be a triangle. We …show more content…
Ex-Touch triangle: If D, E, F are the points of contact of excircles opposite to the vertices A, B, C with the sides BC, CA, AB respectively then the triangle formed by joining the points D, E, F is called as Extouch triangle. It is well known that Ex-Touch triangle is a pedal’s triangle with respect to Bevan point of [7].
5. Bevan point: Bevan point(V) is the Circumcenter of the excentral triangle. It is named in honor of Benjamin Bevan. The Bevan point is the reflection of Incenter in the Circumcenter and it is also the reflection of orthocenter in the Spieker center. The Bevan point(V) is the midpoint of line segment joining the Nagel point and de-Longchamps point.The Bevan point(V) is a triangle center and it is listed as the point X(40) in Clark Kimberling’s Encyclopedia of Triangle Centers[15].
Lemma – 1 If is the area of pedal’s triangle with respect to the any arbitrary point P, and if is the area of reference triangle then prove that Among all the pedal’s triangles, the pedal triangle with respect to some point P whose area is one fourth of the area of the reference triangle, has maximum area. That is . …show more content…
Hence proved.
Lemma – 2
The Ex-Touch triangle of reference triangle acts as the pedal’s triangle with respect to the Bevan point and the area of Ex-Touch triangle is
Proof:
Let D, E, F are the points of tangency of excircles opposite to the vertices A, B, C of with the sides BC, CA, AB and if IA, IB, IC are the excenters opposite to the vertices A, B, C then IAD, IBE, ICF are perpendicular to the sides BC, CA, AB respectively.
Let V is the point of intersection of extensions of IAD, IBE, ICF respectively, now It is enough to prove that the point V is the circumcenter of Excentral triangle IAIBIC that is V is the Bevan point, it results that Ex-Touch triangle is the pedal’s triangle of with respect to Bevan point[1]. Now by angle chasing and using little algebra we can prove that , and
It implies that IAV = IBV = ICV
So V is the Circumcenter of Excentral triangle IAIBIC. Hence Ex-Touch triangle is the pedal’s triangle of with respect to Bevan point.
Now let us find the area of Ex-Touch triangle,
We are familiar with the result that “In any triangle cevian divides the triangle into two triangles whose ratio between the areas is equal to the ratio between corresponding bases”.
So and it implies