Orthocenter & Altitude




Jane Yun



In this exploration, we will construct any inscribed acute triangle, ABC, 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 ABC 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.



Join CP







              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 BHC + Area of  CHA + Area of ABH = Area of ABC.


Area of BHC = ½ (HD * BC)


Area of CHA = ½ (BE * CA)


Area of ABH = ½ (CF * AB)





Since area of ABC = ½ (AD * BC) = ½ (BE * CA) = ½ (CF * AB), area of BHC = ½ (HD * BC),Area of CHA = ½ (BE * CA), and area of ABH = ½ (CF * AB),







Now, previously shown that


and since








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