O is the orthocentre of the `DeltaABC` . Prove that O is also the incentre of its pedal triangles .

Explanation

Let AD, BE and CF be three perpendiculars drawn from the vertices A , B , and C of the ` DeltaABC` to their opposite sides BC, CA and AB repectively which intersect each other at O . Then O is the orthocentre of the `DeltaABC` Let us join D, E, E , F and F, D Then `DeltaDEF` is the pedal triangle of `DeltaABC` .
To prove : O is the incentre of `DeltaDEF`.
Proof : The four points A ,C , D ,F are conoyclic,
[since the two angles `angleDCFandangleDAF` on the same of FD are equal.
`because angleDCF=90^(@)-angleABD=angleBAD=angleFAD]`
Now , the two angles in circle produced by the arc AF are `angleADFandangleACF` ,
`therefore angleADF=angleACF`
`=90^(@)-angleEAF" " [because angleAFC=90^(@),therefore angleACF+angleCAF=90^(@)]`
`=angleABE" " [becauseangleAEB=90^(@),thereforeangleABE+angleEAB=90^(@)]`
= `angleADE` [`because` A ,B D , E are concylic and `angleABE` and
`angleADE ` are two angles in circle produced by arc AE. ]
i .e ., `angleODF=angleODE " " therefore` OD is the bisectore of `angleEDF`
Similarly , it can be proved that OE and OF are the bisectors of `angleDEFandangleDFE` .
So , O lies on the bisectors of the angles of `DeltaDEF` i .e ., O is the point of intersection of the bisectors of the angles of the `DeltaDEF`
Hence O is the incenter of `DeltaDEF`