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In $\triangle ABC$ and construct bisectors of the angles at $A$ and $C$ , intersecting at $O$ 1 Draw $BO$ . We show that $BO$ bisects the angle at $B$ , and that $O$ is in fact the incenter of $\triangle ABC$ .
\begin{pspicture*}(-2.0000,-2.0000)(8.0000,6.0000) \rput(-3,-2){.} \rput(9,6){.} \uput{0.3000}[0.0000](2.5292,1.4169){O} \pspolygon(0.0000,0.0000)(6.0000,0.0000)(2.2500,3.6742) \uput{0.3000}[-180.0000](0.0000,0.0000){A} \uput{0.3000}[0.0000](6.0000,0.0000){B} \uput{0.3000}[90.0000](2.2500,3.6742){C} \psline[linestyle=dotted](-2.0000,-1.1205)(8.0000,4.4819) \psline[linestyle=dotted](8.0000,-0.8165)(-2.0000,3.2660) \psline[linestyle=dotted](2.9519,-2.0000)(1.9623,6.0000) \psarc(0.0000,0.0000){0.5000}{29.2589}{58.5178} \psline(0.3063,0.2946)(0.4144,0.3986) \psarc(0.0000,0.0000){0.5000}{0.0000}{29.2589} \psline(0.4112,0.1073)(0.5564,0.1452) \psdots[dotstyle=*, dotscale=1.2](2.1701,3.3052) \psarc(2.2500,3.6742){0.5000}{-121.4822}{-82.9487} \psdots[dotstyle=*, dotscale=1.2](2.4174,3.3358) \psarc(2.2500,3.6742){0.5000}{-82.9487}{-44.4153} \pscircle(2.5292,1.4169){1.4169} \uput{0.3000}[45.0000](3.5209,2.4290){D} \uput{0.3000}[135.0000](1.3208,2.1569){E}
\uput{0.3000}[-90.0000](2.5292,0.0000){F} \psline(2.5292,1.4169)(3.5209,2.4290) \psline(2.5292,1.4169)(1.3208,2.1569) \psline(2.5292,1.4169)(2.5292,0.0000) \psline(3.6280,2.3241)(3.5230,2.2169)(3.4159,2.3219) \psline(1.3992,2.2848)(1.5271,2.2065)(1.4488,2.0786) \psline(2.3792,0.0000)(2.3792,0.1500)(2.5292,0.1500) \end{pspicture*} \end{center} Drop \htmladdnormallink{perpendiculars}{http://planetmath.org/encyclopedia/PerpendicularPlanes.html} from $O$ to each of the three sides, intersecting the sides in $D$, $E$, and $F$. Clearly, by \htmladdnormallink{AAS}{http://planetmath.org/encyclopedia/AffineCongruence.html}, $\triangle COD \cong \triangle COE$ and also $\triangle AOE\cong\triangle AOF$. Thus $FO=EO=DO$. It follows that $O$ is the incenter of $\triangle ABC$ since its \htmladdnormallink{distance}{http://planetmath.org/encyclopedia/MetricTopology.html} from all three sides is equal. Also, since $FO=DO$ we see that $\triangle BOF$ and $\triangle BOD$ are \htmladdnormallink{right
triangles}{http://planetmath.org/encyclopedia/Triangle.html} with two equal sides, so by \htmladdnormallink{SSA}{http://planetmath.org/encyclopedia/SSA.html} (which is applicable for right triangles), $\triangle BOF\cong\triangle BOD$. Thus $BO$ bisects $\angle ABC$. \end{document}
Footnotes
- 1
- Note that the angle bisectors must intersect by Euclid's Postulate 5, which states that ``if a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles.'' They must meet inside the triangle by considering which side of $AB$ and $CB$ they fall on.
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