1 angle is 60^o when angle of bisectors is 60^o (2009 Euler Olympiad Remote 3.3)

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1 angle is 60^o when angle of bisectors is 60^o (2009 Euler Olympiad Remote 3.3)

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#1 h Feb 5, 2021, 8:24 PM • 2 Y

Y by son7, jasperE3

The two angle bisectors of the triangle intersect at an angle of $60$ degrees. Prove that one of the angles of this triangle is $60$ degrees.

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#2 h Feb 5, 2021, 9:00 PM

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The angle bisector of the angle $\beta$ cuts $AC$ in the point $D$ and cuts the angle bisector of the angle $\gamma$ in the point $E$ .
Given is $\angle CED=60^{\circ}$ .
In $\triangle BCE\ :\ \frac{\beta}{2}+\frac{\gamma}{2}=60^{\circ}$ , so $\beta+\gamma=120^{\circ}$ and $\alpha=60^{\circ}$ .

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#3 h Feb 5, 2021, 9:39 PM • 3 Y

Y by Mango247, Mango247, Mango247

Let us draw a diagram:
$[asy] import olympiad; draw((0,0)--(3,3)); draw((0,0)--(6,0)); draw((6,0)--(3,3)); draw((0,0)--(4.5,1.5)); draw((3,3)--(3,0)); label("A",(0,0),SW); label("B",(3,3),N); label("C",(6,0),SE); label("D",(4.5,1.5),NE); label("E",(3,0),S); label("O",(3,1),NW); [/asy]$

We let $\angle BOD = 60$ . Also we let $\angle EBD = \angle ABE = \alpha$ and $\angle BAD = \angle CAD = \beta$ :

So then we get $\angle BOA = 180 - 60 = 120$ since $A, O, D$ are collinear. Using that, we can now see $\alpha + \beta = 60$ . Now we see that $\angle C = 180 - \angle DAC - \angle ADC = 180 - \alpha - (\angle BOD + \angle OBD) = 180 - \alpha - 60 - \beta = 120 - (\alpha + \beta) = 120 - 60 = 60$ . Therefore, one angle of $\triangle ABC$ is $60^{\circ}$ .

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