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 $AE = EC + CB$ .

#Geometry

Easy Math Editor

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Comments

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.

Vishwash Kumar ΓΞΩ - 4 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$