1. If the points are $P_1$ and $P_2$ and the triangle is $ABC$, Join $AP_1, BP_1$,etc. If $AP=BP=CP$ for any of them that is the circumcenter.(Also remember ABC is scalene) 2. Say $58^{\circ}$ is opposite $AB$. The other angles are $59^{\circ}$ and $63^{\circ}$. But we know that largest side is opposite the largest angle. So $63^{\circ}$ is opposite the larger between $AC$ and $BC$. Peace. Faustus.

Faustus wrote: 1. If the points are $P_1$ and $P_2$ and the triangle is $ABC$, Join $AP_1, BP_1$,etc. If $AP=BP=CP$ 2. So $63^{\circ}$ is opposite the larger between $AC$ and $BC$. Using only the ruler without partitions determine, we can not Compare the length of $AC$ and $BC$.

The three angles are $58^\circ < 59^\circ < 63^\circ $. First we denote ${P_1}{P_2} \cap BC = D$, ${P_1}{P_2} \cap CA = E$, ${P_1}{P_2} \cap AB = F$, Without loss of generality we suppose $D,F$ lie in corresponding segments, then we know $\angle B = 59^\circ $. Without loss of generality we suppose $A$ is between $E,C$, then we know $\angle A = 63^\circ $, and so $\angle C = 58^\circ $. Without loss of generality we suppose ${P_1}$ is between $F,{P_2}$, then we have $\angle A{P_1}B > \angle A{P_2}B$, and $\angle AOB = 2 \times \angle C = 116^\circ $, \[\angle AIB = 90^\circ + \frac{1}{2}\angle C = 119^\circ > \angle AOB\] So we know ${P_1} = I$ and ${P_2} = O$.

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My solution:

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Lawasu wrote: My solution: Using only the ruler without partitions determine , how to "draw $XP \bot AB$"?

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A ruler has a corner, and you can use it.

Lawasu wrote: A ruler has a corner, and you can use it. With a strong ruler, we can do every thing.