Skip to main content
MathOverflow

Return to Question

removed word triangle
Source Link
Pedja
Pedja
  • 2.7k
  • 15
  • 26

Can you provide a proof for the following proposition?

Proposition. Given an arbitrary triangle   $\triangle ABC$. The triangles $\triangle AEB$, $\triangle BFC$ and $\triangle CDA$ are constructed on the sides of the triangle   $\triangle ABC$ either all outward or all inward, such that $\angle DAC=\angle BAE$, $\angle EBA= \angle CBF$, $\angle FCB= \angle ACD$ and $\angle DAC+\angle EBA+\angle FCB=180^{\circ}$. Let $I_1$, $I_2$, $I_3$ be the incenters of the triangles $\triangle AEB$, $\triangle BFC$, $\triangle CDA$, respectively. Then the angles of the triangle   $\triangle I_1I_2I_3$ are equal to $\angle I_3I_1I_2=(\angle BAE+\angle EBA)/2$, $\angle I_1I_2I_3=(\angle CBF+\angle FCB)/2$ and $\angle I_2I_3I_1=(\angle ACD+ \angle DAC)/2$.

Three triangles constructed on the sides of ABC, as in the post body

GeoGebra applet that demonstrates this proposition can be found here.

Can you provide a proof for the following proposition?

Proposition. Given an arbitrary triangle $\triangle ABC$. The triangles $\triangle AEB$, $\triangle BFC$ and $\triangle CDA$ are constructed on the sides of the triangle $\triangle ABC$ either all outward or all inward, such that $\angle DAC=\angle BAE$, $\angle EBA= \angle CBF$, $\angle FCB= \angle ACD$ and $\angle DAC+\angle EBA+\angle FCB=180^{\circ}$. Let $I_1$, $I_2$, $I_3$ be the incenters of the triangles $\triangle AEB$, $\triangle BFC$, $\triangle CDA$, respectively. Then the angles of the triangle $\triangle I_1I_2I_3$ are equal to $\angle I_3I_1I_2=(\angle BAE+\angle EBA)/2$, $\angle I_1I_2I_3=(\angle CBF+\angle FCB)/2$ and $\angle I_2I_3I_1=(\angle ACD+ \angle DAC)/2$.

Three triangles constructed on the sides of ABC, as in the post body

GeoGebra applet that demonstrates this proposition can be found here.

Can you provide a proof for the following proposition?

Proposition. Given an arbitrary   $\triangle ABC$. The $\triangle AEB$, $\triangle BFC$ and $\triangle CDA$ are constructed on the sides of the   $\triangle ABC$ either all outward or all inward, such that $\angle DAC=\angle BAE$, $\angle EBA= \angle CBF$, $\angle FCB= \angle ACD$ and $\angle DAC+\angle EBA+\angle FCB=180^{\circ}$. Let $I_1$, $I_2$, $I_3$ be the incenters of the $\triangle AEB$, $\triangle BFC$, $\triangle CDA$, respectively. Then the angles of the   $\triangle I_1I_2I_3$ are equal to $\angle I_3I_1I_2=(\angle BAE+\angle EBA)/2$, $\angle I_1I_2I_3=(\angle CBF+\angle FCB)/2$ and $\angle I_2I_3I_1=(\angle ACD+ \angle DAC)/2$.

Three triangles constructed on the sides of ABC, as in the post body

GeoGebra applet that demonstrates this proposition can be found here.

(Not) spacing around commas
Source Link
LSpice
LSpice
  • 12.9k
  • 4
  • 45
  • 69

Can you provide a proof for the following proposition:?

Proposition. Given an arbitrary triangle $\triangle ABC$. The triangles $\triangle AEB$  , $\triangle BFC$ and $\triangle CDA$ are constructed on the sides of the triangle $\triangle ABC$ , either either all outward or all inward  , such that $\angle DAC=\angle BAE$  , $\angle EBA= \angle CBF$  , $\angle FCB= \angle ACD$ and $\angle DAC+\angle EBA+\angle FCB=180^{\circ}$  . Let $I_1$,  $I_2$,  $I_3$ be the incenters of the triangles $\triangle AEB$,$\triangle BFC$ $\triangle BFC$,$\triangle CDA$ $\triangle CDA$, respectively. Then the angles of the triangle $\triangle I_1I_2I_3$ are equal to $\angle I_3I_1I_2=(\angle BAE+\angle EBA)/2$  , $\angle I_1I_2I_3=(\angle CBF+\angle FCB)/2$ and $\angle I_2I_3I_1=(\angle ACD+ \angle DAC)/2$  .

enter image description hereThree triangles constructed on the sides of ABC, as in the post body

GeoGebra applet that demonstrates this proposition can be found here.

Can you provide a proof for the following proposition:

Proposition. Given an arbitrary triangle $\triangle ABC$. The triangles $\triangle AEB$  , $\triangle BFC$ and $\triangle CDA$ are constructed on the sides of the triangle $\triangle ABC$ , either all outward or all inward  , such that $\angle DAC=\angle BAE$  , $\angle EBA= \angle CBF$  , $\angle FCB= \angle ACD$ and $\angle DAC+\angle EBA+\angle FCB=180^{\circ}$  . Let $I_1$,$I_2$,$I_3$ be the incenters of the triangles $\triangle AEB$,$\triangle BFC$ ,$\triangle CDA$ , respectively. Then the angles of the triangle $\triangle I_1I_2I_3$ are equal to $\angle I_3I_1I_2=(\angle BAE+\angle EBA)/2$  , $\angle I_1I_2I_3=(\angle CBF+\angle FCB)/2$ and $\angle I_2I_3I_1=(\angle ACD+ \angle DAC)/2$  .

enter image description here

GeoGebra applet that demonstrates this proposition can be found here.

Can you provide a proof for the following proposition?

Proposition. Given an arbitrary triangle $\triangle ABC$. The triangles $\triangle AEB$, $\triangle BFC$ and $\triangle CDA$ are constructed on the sides of the triangle $\triangle ABC$ either all outward or all inward, such that $\angle DAC=\angle BAE$, $\angle EBA= \angle CBF$, $\angle FCB= \angle ACD$ and $\angle DAC+\angle EBA+\angle FCB=180^{\circ}$. Let $I_1$,  $I_2$,  $I_3$ be the incenters of the triangles $\triangle AEB$, $\triangle BFC$, $\triangle CDA$, respectively. Then the angles of the triangle $\triangle I_1I_2I_3$ are equal to $\angle I_3I_1I_2=(\angle BAE+\angle EBA)/2$, $\angle I_1I_2I_3=(\angle CBF+\angle FCB)/2$ and $\angle I_2I_3I_1=(\angle ACD+ \angle DAC)/2$.

Three triangles constructed on the sides of ABC, as in the post body

GeoGebra applet that demonstrates this proposition can be found here.

edited tags
Link
YCor
YCor
  • 63.9k
  • 5
  • 187
  • 285
Source Link
Pedja
Pedja
  • 2.7k
  • 15
  • 26