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)

⇒ ∠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)