Complex numbers - prove |BD| + |CD| = |AD|

by jiten   Last Updated January 16, 2018 16:20 PM

Let $ABC$ be equilateral triangle and $D$ be a point on the arc BC of the circle circumscribing ABC. Then prove using complex algebra that : $|BD| +|CD| = |AD|$.

For me, the confusion is both : (i)- in ability to think of problem, and (ii)- to formulate it.

(i) Ability to think of problem : How is it possible that the two points nearest to $D$ have their arc lengths equal to that of $|AD|$, where A is the distant point. I cannot imagine by taking the shortest length of sum of $|BD| + |CD|$, or taking by a particular direction (cck, ck).

(ii) Unable to find a way to formulate, except possibly by taking the angle possible, and then would involve a range of $60^0$ for any point $D$ on the arc $BC$. Even then difficult to think of about the formulation.

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Let $A(a)$, $B(b)$, $C(c)$ and $D(z)$.

Thus, by Ptolemy $$AB\cdot DC+BD\cdot AC=AD\cdot BC$$ or $$|a-b|\cdot |z-c|+|b-z|\cdot|a-c|=|a-z|\cdot |b-c|$$ and since $$|a-b|=|a-c|=|b-c|,$$ we obtain $$|z-c|+|b-z|=|a-z|$$ or $$CD+BD=AD.$$

Michael Rozenberg
January 16, 2018 16:04 PM

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