Orthocenter & Altitude

 

by

 

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.

 

Q.E.D

 

 

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),

 

then

 

Therefore,

 

 

Now, previously shown that

 

and since

 

therefore,

 

 

Therefore,

 

 


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