# Viviani in Isosceles Triangle: What is it?

A Mathematical Droodle

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Copyright © 1996-2018 Alexander Bogomolny# Viviani in Isosceles Triangle

The applet attempts to illustrate the following proposition:

The sum of distances of a point in the base of an isosceles triangle to the sides does not depend on the position of the point on the base.

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For the proof, let PA'||AC, with A'∈AB. Draw BT'⊥AC and denote S' the intersection of BT' and BS'. Then, since BA'P is isosceles,

PS + PT = BS' + S'T' = BT',

the altitude from B to side AC, which is independent of the position of P.

This statement leads directly to Viviani's theorem:

The sum of distances of a point inside an equilateral triangle or on one of its sides equals the length of its altitude.

With the reference to the diagram below,

Viviani's theorem asserts that the sum PS + PT + PU is indendent of P.

Draw B_{a}C_{a}||BC:

As we just proved, in the isosceles ΔB_{a}AC_{a}, the sum PS + PT is independent of the position of P on B_{a}C_{a}. Since B_{a}C_{a}||BC, this is also true of PU. Therefore, for P∈B_{a}C_{a}, the sum

Since ΔABC is equilateral the same holds for P on A_{b}C_{b}||AC:

It follows that the sum PS + PT + PU is independent of the position of P on the union of the two lines. Now, when P glides over B_{a}C_{a}, A_{b}C_{b} sweeps the whole of ΔABC, showing that the sum remains the same for any position of P inside ΔABC, thus proving Viviani's theorem.

For another proof, simply observe that

Area(ΔABP) + Area(ΔACP) = Area(ΔABC).

Writing that explicitly we obtain

AB·PS/2 + AC·PT/2 = AC·BT'/2,

and, taking into account that AB = AC,

PS + PT = BT' = const.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny63428252 |