×

# 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
10 months, 3 weeks ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears 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} $$\boxed{123}$$

Sort by:

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.

- 10 months, 3 weeks ago