In figure 3.103, seg AD side BC, seg BE side AC, seg CF side AB. Point O is the orthocentre. Prove that , point O is the incentre of∆ DEF.

Join D to E , D to F and E to F.

In ΔABE,

⇒ ∠ ABE + ∠BAE + ∠BEA = 180° (Sum of all angles of a triangle)

⇒ ∠ABE + ∠ BAE + 90 = 180

⇒ ∠ABE + ∠ BAE = 90

⇒ ∠ABO + ∠ BAC = 90

⇒ ∠ABO = 90 - ∠ BAC ………………….(1)

In quadrilateral BFOD, we have

We have ∠F = 90 , ∠D = 90

⇒ ∠B + ∠F + ∠O + ∠D = 360°

⇒ ∠B + ∠O + 180 = 360

⇒ ∠B + ∠O = 180

Therefore, BFOD is a cyclic quadrilateral.

∠FBO = ∠FDO (angle by the same arc)

⇒ ∠ABO = ∠FDO

From (1),

⇒ ∠ FDO = 90 - ∠BAC ………………….(2)

In ΔAFC,

⇒ ∠ CAF + ∠FCA + ∠AFC = 180° (Sum of all angles of a triangle)

⇒ ∠CAF + ∠ FCA + 90 = 180

⇒ ∠CAF + ∠ FCA = 90

⇒ ∠BAC + ∠ OCE = 90

⇒ ∠OCE = 90 - ∠ BAC ………………….(3)

In quadrilateral CEOD, we have

We have ∠E = 90 , ∠D = 90

⇒ ∠C + ∠E + ∠O + ∠D = 360°

⇒ ∠C + ∠O + 180 = 360

⇒ ∠C + ∠O = 180

Therefore, CEOD is a cyclic quadrilateral.

∠ODE = ∠OCE (angle by the same arc)

From (3),

⇒ ∠ ODE = 90 - ∠BAC …………(4)

From (2) and (4) we conclude,

∠FDO = ∠ODE

OD bisects ∠D.

Similarly, we can prove that OE bisects ∠E and OF bisects ∠F.

Hence O is the incenter of ΔDEF.