In the given figure, seg AD ⊥ side BC, seg BE ⊥ side AC, seg CF ⊥ side AB.

Explanation:- 

Given: In triangle ABC, seg AD ⊥ BC, seg BE ⊥ AC and seg CF ⊥ BC. O is the orthocentre of ∆ABC. DE, EF, and DF are joined.

To prove: O is the incentre of ∆DEF.

Proof:

1. We know that AF ⊥ OE and AE ⊥ OF. Therefore, ∠AFO + ∠AEO = 90° + 90° = 180°. Hence, quadrilateral AEOF is cyclic. (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic).
2. Using the property that angles inscribed in the same arc are congruent, we get ∠OAE = ∠OFE (from step 1).
3. We also know that BF ⊥ OD and BD ⊥ OF. Therefore, ∠BFO + ∠BDO = 90° + 90° = 180°. Hence, quadrilateral BFOD is cyclic. (If a pair of opposite angles of a quadrilateral is supplementary, the quadrilateral is cyclic).
4. Again using the property that angles inscribed in the same arc are congruent, we get ∠OBD = ∠OFD (from step 3).
5. In ∆ACD, we know that ∠DAC + ∠ACD = 90° (using angle sum property of a triangle).
6. In ∆BCE, we know that ∠BCE + ∠CBE = 90° (using angle sum property of a triangle).
7. From step 5 and 6, we get ∠DAC + ∠ACD = ∠BCE + ∠CBE.
8. Using step 1, 2, and 7, we get ∠OFE = ∠OFD and ∠OAE = ∠OBD.
9. Therefore, OF is the bisector of ∠EFD, OE is the bisector of ∠DEF, and OD is the bisector of ∠EDF.
10. Hence, O is the incentre of ∆DEF. (Incentre of a triangle is the point of intersection of its angle bisectors).

Therefore, O is the incentre of ∆DEF, which completes the proof.