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Points $A,C_1,B,A_1,C,B_1$ in the indicated order are taken on a circle. Prove that if and only if $AA_1, BB_1, CC_1$ are the bisectors of angles of $ABC$ then they are the heights of $A_1B_1C_1$

This seems little bit confusing to me. I tried drawing a diagram like this -
enter image description here
How $BB_1$ is height of $A_1B_1C_1$. I think it is asking to show that $A_1B_1 \perp CC_1$ or something like that. What am I missed?

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    Okay, draw the segment $A_1C_1$ in GeoGebra and you'll see that it is orthogonal to $BB_1$, this is what you need to prove essentially [ok, you also need to prove converse]. $AB$ and $CC_1$ cannot be orthogonal in the general case, because $CC_1$ is the bisector of $ABC$, and it is a height only if $AC=CB$. – 2017-02-07
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    @Wolfram If I draw $A_1B_1$ then it seems that $CC_1$ is greater than the height of triangle $A_1B_1C_1$ .... – 2017-02-07
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    Ah, you about this. There is some ambiguity in what "height" mean because it can refer to the line segment, length of the line segment or, like here, the full line. Why aren't you surprised that $AA_1$ is greater than the bisector of the triangle $ABC$ here? – 2017-02-07
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    @Wolfram Still didn't understand :'( – 2017-02-07
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    Okay, the definition of the height of $\Delta XYZ$ used here: the height is a **line** (infinite in both directions), containing one of the vertices $X,Y,Z$ and orthogonal to the opposite side. – 2017-02-07
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    Oh :( I was thinking they asked about perpendicular distance to opposite side :| – 2017-02-07

1 Answers 1

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The statement is true and it can be proven easily using angle-chasing. Let $CC_1, BB_1, AA_1$ be angle bisectors in $\triangle ABC$. Also let $CC_1$ intersect $A_1B_1$ at $X$. We'll prove that $\angle C_1XB_1 = \frac{\pi}{2}$.

$$\angle C_1XB_1 = \pi - \angle XB_1C_1 - \angle XC_1B_1 = \pi - \angle XB_1B - \angle BB_1C_1 - \angle XC_1B_1 = \pi - \angle A_1AB - \angle C_1CB - \angle CBB_1 = \pi - \frac{\angle CAB + \angle ACB + \angle ABC}{2} = \pi - \frac{\pi}{2} = \frac{\pi}{2}$$

Therefore $CC_1 \perp A_1B_1$. Similarly $BB_1 \perp A_1C_1$, $AA_1 \perp B_1C_1$. Hence the proof.

For the other side let $CC_1, BB_1, AA_1$ be heighs in $\triangle A_1B_1C_1$. Then we have: $$\angle C_1CA = \angle C_1A_1A = \frac{\pi}{2} - \angle B_1C_1A_1 = \angle BB_1C_1 = \angle BCC_1$$ Therefore $CC_1$ is angle bisector of $\angle ACB$. Similarly we get that $BB_1, AA_1$ are angle bisectors of $\angle ABC$ and $\angle BAC$, respectively. Hence the proof.

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    Yep, this is the solution, simple and clear! Renders all arguments futile :) – 2017-02-08