Altitudes and Orthocenters

By Pei-Chun Shih

The
three altitudes of a triangle intersect in a single point, called the
orthocenter of the triangle. The orthocenter lies inside the triangle if and
only if the triangle is an acute triangle. There are several interesting
relationships between the three altitudes and the orthocenter. Here I am going
to use GSP to explore them and provide proofs of the relationships.

Construct
the orthocenter H in a given acute triangle ABC. Let points D, E, and F be the
feet of the perpendiculars from A, B, and C respectfully. Prove that

+ + = 1 and + + = 2.

The
first proof can be accomplished by using the area formula of a triangle which
is half of the product of a base and its corresponding altitude. LetÕs divide
the triangle ABC into three small triangles: AHC, BHC, and AHB. So the area of
triangle ABC equals the sum triangles AHC, BHC, and AHB:

△AHC + △BHC
+ △AHB = △ABC

Divide
each side by △ABC,

+ + = 1

Express
by the area formula of a triangle,

+ + = 1

Simplify,

+ + = 1 (Proposition
1)

The
second proof can be done by using the proposition 1 we have just proved above.
Since BH = BE – HE, AH = AD – HD, and CH = CF – HF, then we
can do the following substitution:

+ + = + +

Simplify,

= (1 - ) + (1 - ) + (1 - )

Rearrange,

= 3 – ( + + )

Substitute
+ + for 1 and we can
get

+ + = 3 – 1 =
2.

We
can explore the properties of the altitudes and orthocenters further by
constructing the circumcircle of the triangle ABC and extending each altitude
to its intersection with the circumcircle at corresponding points P, Q, and R.

There
is a relationship between the altitudes and AP, BQ, and CR which is + + = 4. We can prove
it by applying the proposition 1 again here. Before using the proposition 1,
letÕs add some construction lines first. Connect AR, BP, and QC. We want to
prove that HD = PD, HE = QE, and HF = RF.

Consider
△BHD and△AHE.
△BHD and△AHE
are similar triangles by the Angle-Angle Similarity Theorem since ∠BHD = ∠AHE
and ∠BDH = 90 degrees = ∠AEH. Therefore, ∠HAE = ∠HBD.
Since points A, E, C are collinear, points A, H, P are collinear, and points B,
D, C are collinear, then ∠HAE
= ∠PAC and ∠PBD
= ∠PBC. ∠PAC
= ∠PBC since they intercept the same arc PC. Hence, ∠PBD = ∠HBD.
△BHD and△BPD
are congruent triangles by the Angle-Side-Angle Congruence Theorem since ∠PBD = ∠HBD,
BD = BD, and ∠BDH = 90 degree = ∠BDP. Therefore, HD = PD. Similarly, we can get that HE
= QE and HF = RF by performing the same steps above for △ARH and△CQH.
Thus, AP = AD + PD = AD + HD, BQ = BE + QE = BE + HE, CR = CF + RF = CF + HF.

So, + + = + +

= (1 + ) + (1 + ) + (1 + )

= 3 + + +

= 3 + 1 by
Proposition 1

= 4

Therefore,
+ + = 4.