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