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Q: The perpendicular bisector of the sides of a triangle AB meet at I. Prove that: IA = IB = IC

enter image description here

By considering two triangles:

We need to prove that $$ I A=I B=I C $$ Proof: In $\Delta$ BID and $\Delta$ CID $\mathrm{BD}=\mathrm{DC}$ (given) $\angle B D I=\angle C D I=90^{\circ}(A D$ is perpendicular bisector of $B C)$ $\mathrm{BC}=\mathrm{BC}$ (common) By SAS postulate of congruent triangles $$ \Delta \mathrm{BID} \cong \triangle \mathrm{CID} $$ The corresponding parts of the congruent triangles are congruent Therefore IB $=I C$ Similarly, In $\Delta \mathrm{CIE}$ and $\triangle \mathrm{AIE}$ $$ \mathrm{CE}=\mathrm{AE} \text { (given) } $$ $\angle \mathrm{CEI}=\angle \mathrm{AEI}=90^{\circ}(\mathrm{AD}$ is perpendicular bisector of $\mathrm{BC})$ IE = IE (common) By SAS postulate of congruent triangles $$ \Delta \mathrm{CIE} \cong \triangle \mathrm{AIE} $$ The corresponding parts of the congruent triangles are congruent Therefore IC = IA Thus, IA = IB = IC

But , I want to know if there is a way I can do this a less lengthier & by considering 3 triangles at once or if considering 3 triangles is even true or possible to do so

Paul Sinclair
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S.M.T
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    The image is very misleading. Perpendicular bisectors in general are not this "neat". For instance, the perpendicular bisector of the side $BC$ does not necessarily intersect $A$; the perpendicular bisector intersects $A$ iff $\angle B = \angle C$. Hence, your statement "$AD$ is perpendicular bisector of $BC$" is generally false. – VTand Nov 26 '21 at 10:02
  • What does it mean by "considering two triangles"? – AltercatingCurrent Nov 26 '21 at 11:43
  • @Mehmer The way I have done by considering two triangles. Proving two triangles congruent & then going onto another two. My Q is can we prove the 3 triangles all at once as congruent ? – S.M.T Nov 26 '21 at 11:44

1 Answers1

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As VTand has pointed out, your diagram is very misleading. Here is a less misleading version:

The common intersection point doesn't even have to lie inside the triangle:

However, one way you can prove it faster is to draw the circle passing through $A,B,C$. The three sides of the triangle are chords of this circle, and are therefore perpendicularly bisected by radii of the circle. That is, the perpendicular bisector of each side must be a radius, and therefore passes through the circle center. Thus the circle center is their common intersection point, and is by its definition equidistant from $A, B, C$.

Whether this truly counts as a shorter demonstration is debatable. After all, it depends on a couple results about circles: that you can find a unique circle through any three non-collinear points in the plane, and that the perpendicular bisectors of chords of a circle are radii.

Paul Sinclair
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