Prove that the altitudes of a triangle are concurrent (meet in one point).
 

ΔABC in which BE and CF are perpendiculars on AC and AB respectively and intersect at O. Join AO and produce it to meet BC on D (fig.)
To Prove: ∠ADC = 90°
Construction: Join EF.
Proof: ∠BEC = ∠BFC = 90°
Points B, C, E, F are concylic       ( ∵ BE and CF are perpendiculars on AC and AB respectively and intersect at O)
∠ACD + ∠BFE = 180°
⇒ ∠ACD + 90° + ∠OFE = 180°
⇒ ∠ACD + ∠OFE = 90°       .................... (i)
Now ∠OFA + ∠OEA = 180°  .................... (ii)
Hence, A, F, O, E are concylic.
∠OFE = ∠OAE (angles in the same segment of circle)
From Equations (i) and (ii) we get
∠ACD + ∠OAE = 90°
∠ACD + ∠DAC = 90°
But ∠ADC + (∠ACD + ∠DAC) = 180°
⇒ ∠ADC + 90° = 180°
⇒ ADC = 90°