**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,

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),

then

Therefore,

Now,
previously shown that

and
since

therefore,

Therefore,