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 \(AE = EC + CB\).

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 \(AE = EC + CB\)....

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TopNewestIt 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 \(AE = EC + CB\)....

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