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In the adjoining figure, 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.

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Given: seg AD ⊥ side BC, seg BE ⊥ side AC, 

seg CF ⊥ side AB.

To prove: Point O is the incentre of ∆DEF. 

Construction: Draw DE, EF and DF. 

Proof: ∠OFA = ∠OEA = 90° [Given]

Now, ∠OFA + ∠OEA = 90° + 90° 

∴ ∠OFA + ∠OEA = 180° 

∴ □OFAE is a cyclic quadrilateral. [Converse of cyclic quadrilateral theorem] 

∴ Points O, F, A, E are concyclic points. 

∴ seg 0E subtends equal angles ∠OFE and ∠OAE on the same side of OE. 

∴ ∠OFE = ∠OAE ……… (i) ∠OFB ∠ODB = 90° [Given] 

Now, ∠OFB + ∠ODB = 90° + 90° 

∴ ∠OFB + ∠ODB = 180° 

∴ OFBD is a cyclic quadrilateral [Converse of cyclic quadrilateral theorem] 

∴ Points O, F, B, D are concyclic points

∴ seg OD subtends equal angles ∠OFD and 

∠OBD on the same side of OD. 

∠OFD = ∠OBD ………….. (ii) 

In ∆AEO and ∆BDO, 

∠AEO = ∠BDO [Each angle is 90°] 

∠AOE = ∠BOD [Vertically opposite angles] 

∴ ∆AEO ~ ∆BDO [AA test of similarity] 

∴ ∠OAE = ∠OBD …………….. (iii) [Corresponding angles of similar triangles] 

∴ ∠OFE = ∠OFD [From (i), (ii) and (iii)] 

∴ ray FO bisects ∠EFD. 

Similarly, we can prove ray EO and ray DO bisects ∠FED and ∠FDE respectively. 

∴ Point O is the intersection of angle bisectors of ∠D, ∠E and ∠F of ∆DEF. 

∴ Point O is the incentre of ∆DEF.

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