Pedal Triangle of a Triangle: Let Delta ABC be any triangle and let D, E, F be the feet of perpendiculars from the vertices A, B, C on the opposite sides BC, CA, AB respectively, then the triangle DEF is known as Pedal triangle of ABC. H is the orthocentre of the Delta ABC. We note that angle HDC= angle HEC=90^(@)' therefore angle HDE= angle HCE =90^(@) -A. Simillarly angle HDF =angle HBF=90^(@)-A. Hence, angle FDE=180^(@)-2A. Identically we can find angle DEF = 180^(@)-2B and angle EFD=180^(@) -2C. Thus, the angles of Pedal triangle are 180^(@)-2A, 180^(@) -2B, 180^(@)-2C. Further in Delta BFD, angle FDB=90^(@)-(90^(@)-A) =A, thus (FB)/(sin B)=(FB)/(sin A) therefore FD=(sin B)/(sin A) (BC cos B) = (a sin B cos B)/(sin A) = 2R sin B cos B=b cos B. Similarly, EF=a cos A and DE =c cos C. Thus, the sides of pedal triangle are a cos A, b cos B and c cos C (or R sin 2A, R sin 2B, R sin 2C), where R is the circumradius of Delta ABC. In the above derivation, the triangle ABC is assumed an acute angled triangle. In case the triangle ABC be obtuse angled with A as obtuse angle, then, the angles of pedal triangle will be represented by 2A - 180^(@), 2B,2C andthe sides will be represented by -a cos A, b cos B, c cos C. The inradius of the pedal triangle of a tringle ABC is