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

Note by Tristan Shin 4 years, 3 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.

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## Comments

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TopNewestI 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.

Log in to reply