In an isosceles triangle $ABC$ show that $PM+PN$ does not depend on the position of the chosen point P.

Observe that $$\frac{AB·PM}{2}+\frac{AC·PN}{2}=\frac{BC·h_a}{2}=\text{Area of }\Delta ABC$$ Since $AB=AC$

$$PM+PN=\frac{BC·h_a}{AB}=h_b=h_c$$

which is constant (and therefore doesn't depend on $P$'s position). $\;\blacksquare$

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Geometry

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