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Law of Altitudes

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

Note by Tristan Shin
2 years, 11 months ago

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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. · 2 years, 11 months ago