I is the incenter of triangle ABC with
B' the intersection of the ray BI and
side AC and C' the intersection of ray
CI and side AB.
Prove that AB'IC' is a cyclic quadrilateral if and only if angle BAC is 60 degrees.
The incenter implies the rays are angle bisectors.
A quadrilateral is cyclic if and only if two opposite angles are supplementary.
Let the angles of the triangle be a,b,c.
It's simple to find that angle IB'A=b/2+c and angle IC'A=b+c/2
These angles are opposite angles of the quadrilateral, so the quadrilateral is cyclic if and only if (b/2+c)+(b+c/2)=180
3b/2 + 3c/2 = 180
b+c=120
a=60
----------------------
Playing with Geometer's Sketchpad I found an interesting connection among the other two angles of the quadrilateral:
angle B'IC' = 90 + a/2
This is a little harder to prove, but can also be used to show a=60.
|
|
Posted by Jer
on 2017-08-27 17:58:50 |
Solution & a further result