Orthocenter & Altitude
In this exploration, we will construct any inscribed acute triangle, ÆABC, in the circle, draw three altitudes AD, BE, and CF, and extend AD, BE, and CF toward P, Q, & R so that it will intersect the circle. Then, we want to find
First, we are going to prove that HD = DP, HE = EQ, and HF = FR.
Proof: Let ÆABC be any acute triangle that is inscribed in the circle and let AD, BE, and CF be the altitudes.
Extend AD, BE, and CF toward P, Q, & R so that it intersects the circle.
Want to prove that HD = DP.
Therefore, HD = DP.
Similarly, we can prove that the HE = EQ and HF = FR.
Now, we want to find
By looking at the construction,
AP = AD + DP
BQ = BE + EQ
CR = CF + FR
Then, by substitution
Since HD = DP, HE = EQ, and HF = FR,
this is equivalent to
Area of ÆBHC + Area of ÆCHA + Area of ÆABH = Area of ÆABC.
Area of ÆBHC = ½ (HD * BC)
Area of ÆCHA = ½ (BE * CA)
Area of ÆABH = ½ (CF * AB)
Since area of ÆABC = ½ (AD * BC) = ½ (BE * CA) = ½ (CF * AB), area of ÆBHC = ½ (HD * BC),Area of ÆCHA = ½ (BE * CA), and area of ÆABH = ½ (CF * AB),
Now, previously shown that