AD, BE and CF are the perpendiculars from the | vedclass

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(C) The triangle $\Delta DEF$ is the orthic triangle of $\Delta ABC$. The sides of the orthic triangle are $a \cos A, b \cos B, c \cos C$.Using the si

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$\frac{r}{R}$

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(C) The triangle $\Delta DEF$ is the orthic triangle of $\Delta ABC$. The sides of the orthic triangle are $a \cos A, b \cos B, c \cos C$.Using the sine rule,$a = 2R \sin A, b = 2R \sin B, c = 2R \sin C$.The perimeter of $\Delta DEF$ is $P_{DEF} = 2R \sin A \cos A + 2R \sin B \cos B + 2R \sin C \cos C = R(\sin 2A + \sin 2B + \sin 2C)$.Using the identity $\sin 2A + \sin 2B + \sin 2C = 4 \sin A \sin B \sin C$,we get $P_{DEF} = 4R \sin A \sin B \sin C$.The perimeter of $\Delta ABC$ is $P_{ABC} = a + b + c = 2R(\sin A + \sin B + \sin C)$.Using the identity $\sin A + \sin B + \sin C = 4 \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$,we have $P_{ABC} = 8R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$.Also,$r = 4R \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.Ratio $= \frac{4R \sin A \sin B \sin C}{2R(\sin A + \sin B + \sin C)} = \frac{4R (8 \sin \frac{A}{2} \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{B}{2} \sin \frac{C}{2} \cos \frac{C}{2})}{8R \cos \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}} = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2} = \frac{r}{R}$.

$AD, BE$ and $CF$ are the perpendiculars from the vertices of a $\Delta ABC$ to the opposite sides. The ratio of the perimeter of $\Delta DEF$ to the perimeter of $\Delta ABC$ is: (where $r$ is the inradius and $R$ is the circumradius of $\Delta ABC$)

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