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 triangleâ€™s vertices to their respective opposite sides.

**Pedal triangle Properties**

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

In right angle âˆ†AEB, âˆ AEB=90Â°

Cos A= AE/AB= AE/c

**AE=c.Cos A**

Similarly in âˆ†AFC, **AF=b.Cos A**

Use the ** Cosine formula** in âˆ†ABC,

BCÂ²=ABÂ²+ACÂ²-2Ã—ABÃ—ACÃ—Cos A

**aÂ²=bÂ²+cÂ²-2bc.Cos A**

Similarly in âˆ†AFE,

FEÂ²=AFÂ²+AEÂ² â€“ 2.AF.AE.Cos A

FEÂ²=bÂ² CosÂ²A+cÂ² CosÂ²A â€“ 2bc CosÂ³A

=CosÂ²A (bÂ²+cÂ² â€“ 2bc Cos A)

= aÂ² CosÂ²A

**FE=a.Cos A**

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

**Interior Angles of Orthic Triangle**

BFOD is a ** Cyclic Quadrilateral** (âˆ BFO+âˆ BDO=180Â°)

In right angle âˆ†ABE, âˆ AEB=90Â°

So, âˆ ABE=90Â°-A

For chord FO, **âˆ FBO=âˆ FDO=90Â°-A**

Similarly, for ** Cyclic Quadrilateral **DOEC,

**âˆ ECO=âˆ EDO=90Â°-A**

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

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

âˆ FDE=âˆ FDO+âˆ EDO=180Â°-2A=Ï€-2A

**âˆ FDE=Ï€-2A**

**âˆ DEF=Ï€-2B**

**âˆ DFE=Ï€-2C**

**Area of Orthic Triangle**

Area of âˆ†DEF=1/2Ã—DFÃ—DEÃ—Sin (âˆ FDE)

=1/2Ã—b.Cos BÃ— c.Cos CÃ—Sin(180Â°-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 âˆ†ABC. (2.Cos A.Cos B.Cos C)

ar(âˆ†DEF)/ar(âˆ†ABC)=2.Cos A.Cos B.Cos C

**Circumradius of Orthic Triangle**

In the given figure, **R **is the Circumradius of âˆ†ABC and **X **is the Circumradius of âˆ†DEF.

Apply * Sine Rule* in âˆ†ABC

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

Similarly for âˆ†DEF, \frac{FE}{Sin(âˆ FDE)}=2X.

FE=2X. Sin(180Â°-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 âˆ†DEF=p(âˆ†DEF) and Perimeter of âˆ†ABC=p(âˆ†ABC)

p(âˆ†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(180Â°-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 {180Â°-(A+B)}]

=2R. Sin C {Cos (A-B) â€“ 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}{2RÂ²}. â€¦â€¦..(R=abc/4âˆ†), Area of the triangle=âˆ†

=\frac{(R.4âˆ†)}{2RÂ²}.

=\frac{2âˆ†}{R}.

=\frac{(2. rs)}{R}. â€¦â€¦..(âˆ†=rs), Inradius of the triangle=r, Semiperimeter=s

=\frac{r.(2s)}{R}. â€¦â€¦..(Perimeter of the triangle=2s)

=\frac{r. p(âˆ†ABC)}{R}.

p(âˆ†DEF) : p(âˆ†ABC)= r : R

Hence Provedâ€¦

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