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Pedal Triangle Properties, theorems and their applications in Geometry. Geometry shortcuts and formulas with proof are given below:

Pedal Triangle: The triangle, whose vertices are the feet of the perpendiculars drawn from an arbitrary point inside the triangle to the sides of the triangle.

Orthic Triangle: The triangle, formed by joining the feet of the altitudes drawn from the triangle’s vertices to their respective opposite sides.

Pedal triangle Properties

In the given ∆ABC, O is the intersection point of the altitudes of the triangle (Orthocenter of ∆ABC).

In right angle ∆AEB, ∠AEB=90°

Cos A= AE/AB= AE/c

AE=c.Cos A

Similarly in ∆AFC, AF=b.Cos A

Use the Cosine formula in ∆ABC,

BC²=AB²+AC²-2×AB×AC×Cos A

a²=b²+c²-2bc.Cos A

Similarly in ∆AFE,

FE²=AF²+AE² – 2.AF.AE.Cos A

FE²=b² Cos²A+c² Cos²A – 2bc Cos³A

=Cos²A (b²+c² – 2bc Cos A)

= a² Cos²A

FE=a.Cos A

Similarly, DF=b. Cos B  and ED=c.Cos C

Interior Angles of Orthic Triangle

BFOD is a Cyclic Quadrilateral (∠BFO+∠BDO=180°)

In right angle ∆ABE, ∠AEB=90°

So, ∠ABE=90°-A

For chord FO, ∠FBO=∠FDO=90°-A

Similarly, for Cyclic Quadrilateral DOEC, ∠ECO=∠EDO=90°-A

From both equations: ∠FDO=∠EDO=90°-A (OD is the Angle Bisector)

Now we can say that point O is the Incenter of ∆DEF (Intersection point of angle bisectors DO, EO & FO). Whereas O is also the Orthocenter of ∆ABC.

∠FDE=∠FDO+∠EDO=180°-2A=π-2A

∠FDE=π-2A

∠DEF=π-2B

∠DFE=π-2C

Area of Orthic Triangle

Area of ∆DEF=1/2×DF×DE×Sin (∠FDE)

=1/2×b.Cos B× c.Cos C×Sin(180°-2A)

=1/2.bc.Cos B.Cos C. Sin 2A

=1/2.bc.Cos B.Cos C. (2. Sin A.Cos A)

=1/2.bc Sin A (2. Cos A.Cos B.Cos C)

=Area of ∆ABC. (2.Cos A.Cos B.Cos C)

ar(∆DEF)/ar(∆ABC)=2.Cos A.Cos B.Cos C

Circumradius of Orthic Triangle

In the given figure, R is the Circumradius of ∆ABC and X is the Circumradius of ∆DEF.

Apply Sine Rule in ∆ABC

$\frac{a}{Sin A}=\frac{b}{Sin B}=\frac{c}{Sin C}=2R$.

Similarly for ∆DEF, $\frac{FE}{Sin(∠FDE)}=2X$.

FE=2X. Sin(180°-2A)

a.Cos A=2X. Sin 2A

a.Cos A=2X. (2. Sin A.Cos A)

a/2.Sin A=2X

R=2X

The perimeter of the Orthic Triangle

The perimeter of ∆DEF=p(∆DEF)   and Perimeter of ∆ABC=p(∆ABC)

p(∆DEF)=FE+DF+DE

=a.Cos A+b.Cos B+c.Cos C

=2R.Sin A.Cos A+2R.Sin B.Cos B+2R.Sin C.Cos C

=R (Sin 2A+Sin 2B+Sin 2C)

=R {2.Sin(A+B). Cos(A-B)+Sin 2C}

=R {2.Sin(180°-C). Cos(A-B)+Sin 2C}

=R {2.Sin C. Cos(A-B)+ 2.Sin C.Cos C}

=2R.Sin C {Cos(A-B) +Cos C}

=2R. Sin C [Cos(A-B) +Cos {180°-(A+B)}]

=2R. Sin C {Cos (A-B) – Cos (A+B)}

=2R. Sin C (2.Sin A. Sin B)

=4R. Sin A. Sin B. Sin C

=4R. (a/2R). (b/2R). (c/2R)

=$\frac{abc}{2R²}$.                      ……..(R=abc/4∆), Area of the triangle=∆

=$\frac{(R.4∆)}{2R²}$.

=$\frac{2∆}{R}$.

=$\frac{(2. rs)}{R}$.                   ……..(∆=rs), Inradius of the triangle=r, Semiperimeter=s

=$\frac{r.(2s)}{R}$.                  ……..(Perimeter of the triangle=2s)

=$\frac{r. p(∆ABC)}{R}$.

p(DEF) : p(ABC)= r : R

Hence Proved…

The Centroid of the Triangle (All Concepts): Click Here