Orthocenter & Altitude

by

Jane Yun

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.

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,

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),

then

Therefore,

Now, previously shown that

and since

therefore,

Therefore,