Angles in a triangle

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Angles in a triangle

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Bugi

#1 h Nov 12, 2008, 2:13 PM • 2 Y

Y by Adventure10, Mango247

In triangle $ABC,H,I,O$ are orthocenter, incenter and circumcenter, respectively. $CI$ cuts circumcircle at $L$ . If $AB=IL$ and $AH=OH$ , find angles of triangle $ABC$ .

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#2 h Nov 12, 2008, 4:12 PM • 2 Y

Y by Adventure10, Mango247

Is the answer 120, 15, 45?

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Bugi

#3 h Nov 12, 2008, 7:20 PM • 2 Y

Y by Adventure10, Mango247

Please elaborate on how did you get the solution.

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#4 h Nov 12, 2008, 8:18 PM • 3 Y

Y by Adventure10, Math_legendno12, Mango247

It's easy to prove that BCL is equilateral (use lemma - BL=IL=CL). Let AC, BH intersect at P, AC, AB - at Q. We have $\angle QAC=\angle PAB=60^{\circ}$ => $\angle HCP=\angle HPQ=30^{\circ}$ and $\angle BHC=60^{\circ}$ . Hence HCOB is inscribed quadrilateral, and HO is bisector of CHB. Now using AH=OH and $\angle OHB=30^{\circ}$ we get $\angle HBO=75^{\circ}$ => $\angle ABC=15^{\circ}$ (because <HBQ= <CBO=30).

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#5 h Jun 5, 2021, 4:27 AM

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Angles $\angle LAB$ and $\angle LCB$ share the same arc, so $\angle LAB = \frac{1}{2} \angle C$ . Similarly, $\angle LBA = \frac{1}{2} \angle C$ . Using the same idea $\angle BLC = \angle A$ and $\angle ALC = \angle B$ . Since $\angle LBI = \frac{1}{2} (\angle B + \angle C)$ , then $\triangle LBI$ dictates that $\angle LIB = \angle LBI$ and so $IL=AL=BL=AB$ . Or you may have already known this result.

Since $\triangle ABL$ is equilateral, $\angle A + \angle B = 60^{\circ}$ and $\angle C = 120^{\circ}$ . From observing the triangle formed by $A$ , $B$ , and the foot of $B$ onto $AH$ , $\angle BAH = 90^{\circ} - \angle B$ . Likewise, $\angle ABH = \angle 90^{\circ}-\angle A$ , so $\angle AHB = \angle A + \angle B = 60^{\circ}$ . From $AH=OH$ , $\angle AOH = \angle OAH = 120^{\circ} - \angle B$ , so $\angle AHO = 2\angle B - 60^{\circ}$ . Since $\angle AOB = 2\angle ALB = 120^{\circ}$ , $AOBH$ is cyclic, and so $\angle AHO = \angle ABO \Rightarrow 2 \angle B - 60^{\circ} = 30^{\circ} \Rightarrow \angle B = 45^{\circ}, \angle A = 15^{\circ}$ .

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