Question

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

Answer 1

It is given that seg AD ⊥ side BC, seg BE ⊥ side AC and seg CF ⊥ side BC. O is the orthocentre of ∆ABC.
Join DE, EF and DF.

∠AFO + ∠AEO = 90º + 90º = 180º 
∴ Quadrilateral AEOF is cyclic.     (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic)
⇒ ∠OAE = ∠OFE          .....(1)       (Angles inscribed in the same arc are congruent)
∠BFO + ∠BDO = 90º + 90º = 180º 
∴ Quadrilateral BFOD is cyclic.     (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic)

∴ seg OD subtends equal angles ∠OFD and ∠OBD on the same side of OD.
⇒ ∠OBD = ∠OFD         .....(2)       (Angles inscribed in the same arc are congruent)
In ∆ACD,
∠DAC + ∠ACD = 90º   .....(3)        (Using angle sum property of triangle)
In ∆BCE,
​∠BCE + ∠CBE = 90º    .....(4)        (Using angle sum property of triangle)
From (3) and (4), we get
∠DAC + ∠ACD = ∠BCE + ∠CBE
⇒ ∠DAC = ∠CBE       .....(5)
From (1), (2) and (5), we get
∠OFE = ∠OFD
⇒ OF is the bisector of ∠EFD.
Similarly, OE and OD are the bisectors of ∠DEF and ∠EDF, respectively.
Hence, O is the incentre of ∆DEF.        (Incentre of a triangle is the point of intersection of its angle bisectors)