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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°
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.
