# Prove it

Let $$ABC$$ be a triangle with $$AC>BC$$. Let $$D$$ be the midpoint of the arc $$AB$$ that contains $$C$$, on the circumcircle of $$\bigtriangleup ABC$$. Let $$E$$ be the foot of the perpendicular from $$D$$ on $$AC$$. Prove that $$AE = EC + CB$$.

Note by Mithil Shah
1 year, 3 months ago

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## Comments

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It is the Archimedes Broken Chord Theorem, the solution directly follows.

Here is my proof for the following theorem. Note : - It is a proof for the theorem so, I have proved that $$AD$$ = $$BC$$ + $$DC$$. Using this theorem you can get your desired result. As in the setup given in Prove It is $$D$$ is the midpoint of the arc $$ACB$$ and also, $$DE \perp AC$$ .Therefore $$AE = EC + CB$$....

K.I.P.K.I.G.

- 1 year, 3 months ago

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