0
$\begingroup$

In the given figure, $O$ is the center of the circle. If $AC$ is the bisector of $\angle OAB$, then prove that $AB=BC$.

enter image description here

My attempt, $\angle OAC=\angle CAB $ (AC is the bisector)

$OA=OB$ $\angle OAB=\angle OBA$.

  • 0
    Are OA and BC parallel?? – 2017-01-17
  • 0
    @Rohan, No, they are not. – 2017-01-17
  • 1
    @NeWtoN -- You need more information. With the information you gave, it's not always true. For example, if $\angle AOB$ is close to zero, then $BC$ will be much larger than $AB$. – 2017-01-17
  • 0
    @quasi The statement is true iff $\triangle OAB$ is equilateral. Making $OA$ parallel to $BC$ also works. – 2017-01-17
  • 0
    @Arthur -- Yes, I was about to point that out. So, most likely, NeWtoN's posted version of the problem omitted some information that was present in the actual problem. – 2017-01-17

1 Answers 1

0

See, that the triangle $AOB$ is isosceles.

Let $\angle AOB = 2\alpha$, $\angle OAB = 2\beta$. Then:

  • $\angle OBA = 2\beta$
  • $\angle OAC = \beta$
  • $\angle BAC = \beta$
  • $\angle AOC = \alpha$

We have then: $$\angle AOB + \angle ABO+ \angle OAB =180^{\circ}$$ $$2\alpha +2\beta + 2\beta =180^{\circ}$$

$$\alpha +2\beta=90^{\circ}$$

We know, that $AB=BC$ iff triangle $ABC$ is isosceles and $\angle BAC = \angle ACB$. So we have then $\alpha = \beta = 30^{\circ}$

So $AB=BC$ iff triangle $OAB$ is equilateral.