# 90 is not equal 90

Given that $$AB=DC,DA$$perpendicular to $$AB$$ . Let M and N be the perpendicular bisector of $$AD$$ and $$BC$$ respectively. These two perpendicular bisector intersect in $$E$$. Joining $$AE,ED,EB,EC$$ since $$EM$$ is perpendicular bisector $$AE=ED$$ since $$EN$$ is perpendicular bisector $$EB=EC$$ $$\triangle EAB=\triangle EDC$$ (SSS) $$\angle EAB=\angle EDC$$ $$\angle EAD=\angle EDA$$ we have $$\angle MAB=\angle MDC$$ obviously it is impossible.

Note by Choi Chakfung
2 years, 2 months 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:

Here's an accurate graphic of how it should look like

We can see that $$\Delta ABE$$ is congruent with $$\Delta DCE$$, and the paradox then vanishes.

- 2 years, 1 month ago