×

Law of Altitudes

In triangle $$ABC$$ with altitudes $$AD$$, $$BE$$, and $$CF$$, prove that $$ADsin(A) = BEsin(B) = CFsin(C)$$.

Note by Tristan Shin
3 years, 10 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:

I did it.

Let $$AB=c, AC=b, BC=a$$.

Then

$$A=\frac { aAD }{ 2 } =\frac { bBE }{ 2 } =\frac { cCF }{ 2 }$$,

which gives us the formulas for the altitudes:

$$AD=\frac { 2A }{ a } ,\quad BE=\frac { 2A }{ b } ,\quad CF=\frac { 2A }{ c }$$

For the sines we will use the formulas:

$$A=\frac { bcsin(A) }{ 2 } =\frac { acsin(B) }{ 2 } =\frac { absin(C) }{ 2 }$$,

which gives us:

$$sin(A)=\frac { 2A }{ bc } ,\quad sin(B)=\frac { 2A }{ ac } ,\quad sin(C)=\frac { 2A }{ ab }$$

If we plug everything in the relation we have to prove we get:

$$\frac { 4{ A }^{ 2 } }{ abc } =\frac { 4{ A }^{ 2 } }{ abc } =\frac { 4{ A }^{ 2 } }{ abc }$$

which is true.

- 3 years, 10 months ago