Pedal Triangle

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.

Pedal Triangle Properties
Pedal Triangle Properties

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

Orthic Triangle
Orthic Triangle

Pedal triangle Properties

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

Length of the sides of triangle
The Pedal triangle Properties

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

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Interior Angles of Orthic Triangle

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

In right angle ∆ABE, ∠AEB=90°

So, ∠ABE=90°-A

Interior Angles
Pedal triangle Properties

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.

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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) 

Area of Orthic Triangle
The Pedal triangle and Orthic Triangle

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.

Circumradius of Orthic Triangle
Circumradius of Orthic Triangle

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


The perimeter of the Orthic Triangle

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

Area of Orthic Triangle
Perimeter of Orthic Triangle


=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{(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

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