In figure 3.103, 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 ΔDEF.

Construction: Draw seg DE, seg EF and seg FD.

Proof:

In \(\square BDOF\), \(\angle BDO = \angle BFO = 90^\circ\)

\(\therefore \angle BDO + \angle BFO = 180^\circ\), so \(\square BDOF\) is cyclic (opposite angles supplementary).

\(\therefore \angle ODF = \angle OBF\) (angles inscribed in the same arc OF)

i.e. \(\angle ADF = \angle EBF\) ...(1)

Similarly, in \(\square CDOE\), \(\angle CDO = \angle CEO = 90^\circ\), so \(\square CDOE\) is cyclic.

\(\therefore \angle ODE = \angle OCE\) (angles inscribed in the same arc OE)

i.e. \(\angle ADE = \angle FCE\) ...(2)

Now \(\angle BFC = \angle BEC = 90^\circ\), so seg BC subtends equal angles at F and E, which lie on the same side of BC. \(\therefore\) B, C, E, F are concyclic.

\(\therefore \angle EBF = \angle FCE\) (angles inscribed in the same arc EF) ...(3)

From (1), (2) and (3), \(\angle ADF = \angle ADE\).

\(\therefore\) seg DO (ray DA) bisects \(\angle EDF\).

In the same way we can prove that seg EO bisects \(\angle DEF\) and seg FO bisects \(\angle DFE\).

Thus the angle bisectors of \(\Delta DEF\) pass through O, and the point of concurrence of angle bisectors is the incentre.

\(\therefore\) O is the incentre of \(\Delta DEF\).