Law of Altitudes

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

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Comments

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.

Adrian Neacșu - 7 years, 1 month 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}`	$\boxed{123}$