Tenth Grade Mathematics: Apollonius' Theorem, Stewart's Theorem, and Angle Bisectors

Beyond the standard curriculum of right-angled similarity lies a rich collection of advanced Euclidean theorems that govern arbitrary oblique triangles. When medians, angle bisectors, or general cevianas are drawn inside a triangle, their lengths and segment divisions satisfy precise algebraic relations. Apollonius' Theorem relates the lengths of the sides of a triangle to the length of its median. Stewart's Theorem generalizes this to any internal transversal segment. The Angle Bisector Theorem links opposite side divisions directly to the ratio of adjacent sides.

The Angle Bisector Theorem and its Converse

Theorem (Internal Angle Bisector Theorem):
The internal bisector of an angle of a triangle divides the opposite side internally
in the ratio of the sides containing the angle.

Geometric Diagram:
                         A
                        /|                       / |                       /  |  \   AD is the bisector of Angle A
                     /   |   \  Angle BAD = Angle CAD
                  B +----+----+ C
                         D
               Condition: BD / DC = AB / AC

Proof of the Angle Bisector Theorem

Given: In triangle ABC, AD is the internal bisector of Angle BAC, meeting BC at D. To Prove: BD / DC = AB / AC. Construction: Through vertex C, draw a line parallel to DA to meet the extended ray of BA at point E (CE || DA).

Proof: Since CE || DA and AC is a transversal: Angle CAD = Angle ACE (alternate interior angles). Since CE || DA and BE is a transversal: Angle BAD = Angle AEC (corresponding angles). Since AD bisects Angle BAC: Angle BAD = Angle CAD. Therefore: Angle ACE = Angle AEC. In triangle ACE, two angles are equal, so the sides opposite them are equal:

AE = AC     --- (Equation 1)

Now, in triangle BCE, DA is parallel to CE (DA || CE). By the Basic Proportionality Theorem:

BD / DC = BA / AE

Substituting AE = AC from Equation 1:

BD / DC = AB / AC

This completes the proof.

Converse of the Angle Bisector Theorem:
If a line segment from a vertex to the opposite side divides that side in the
ratio of the remaining two sides, then the segment bisects the vertex angle.

Apollonius' Theorem

Apollonius' Theorem connects the length of any median of a triangle to the lengths of its three sides.

Theorem (Apollonius' Theorem):
In any triangle ABC, if AD is a median bisecting side BC (so BD = DC), then:
                   AB^2 + AC^2 = 2 * (AD^2 + BD^2)

Derivation via Pythagoras' Theorem:
Draw altitude AE perpendicular to BC.
Case 1: In right triangle ABE: AB^2 = AE^2 + BE^2.
Case 2: In right triangle ACE: AC^2 = AE^2 + CE^2.
Notice that in right triangle ADE: AE^2 = AD^2 - DE^2.
Express BE = BD - DE and CE = CD + DE = BD + DE (since BD = CD).
Expanding the squares:
AB^2 + AC^2 = 2*AE^2 + (BD - DE)^2 + (BD + DE)^2
            = 2*(AD^2 - DE^2) + 2*BD^2 + 2*DE^2
            = 2*AD^2 - 2*DE^2 + 2*BD^2 + 2*DE^2
            = 2 * (AD^2 + BD^2).

Stewart's Theorem: The Master Cevian Relation

A cevian is any line segment joining a vertex of a triangle to an arbitrary point on the opposite side. Stewart's Theorem, published by Scottish mathematician Matthew Stewart in 1746, provides the universal length formula for any cevian.

Theorem (Stewart's Theorem):
Let d be a cevian of length d drawn from vertex A to side BC of triangle ABC,
dividing BC of length a into segments of lengths m and n (such that m + n = a):

                     b^2 * m + c^2 * n = a * (d^2 + m * n)

Visual Mnemonic ("A Man and His Boat"):
                 A
                /|            c  / | \  b
              /  |d            B +---+---+ C
               m   n
             <--- a --->
Mnemonic: b*m*b + c*n*c = a*(d*d + m*n)  ===>  "bmb + cnc = man + dad"

Notice that when the cevian is a median, m = n = a/2. Substituting m = n = a/2 into Stewart's Theorem yields Apollonius' Theorem directly!

Chapter Exercises

Exercise 1: In triangle ABC, AD is the internal bisector of Angle A. If AB = 5.6 cm, AC = 6 cm, and DC = 3 cm, find the length of BC.

Exercise 2: The sides of a triangle ABC are AB = 7 cm, BC = 8 cm, and AC = 9 cm. Find the length of the median drawn from vertex A to side BC using Apollonius' Theorem.

Exercise 3: In triangle ABC, AB = 5 cm, AC = 7 cm, and BC = 6 cm. A point D on BC divides BC in the ratio 2 : 1 (so BD = 4 cm and DC = 2 cm). Use Stewart's Theorem to find the exact length of the cevian AD.

Solutions and Explanations

Solution 1: We are given that AD bisects Angle A in triangle ABC. By the Internal Angle Bisector Theorem: BD / DC = AB / AC. Substituting the known values: BD / 3 = 5.6 / 6. Multiply both sides by 3: BD = (5.6 * 3) / 6 = 5.6 / 2 = 2.8 cm. The total length of side BC is BC = BD + DC = 2.8 + 3 = 5.8 cm.

Solution 2: Let AD be the median drawn to side BC. Then D is the midpoint of BC, so BD = DC = BC/2 = 8/2 = 4 cm. By Apollonius' Theorem: AB^2 + AC^2 = 2 * (AD^2 + BD^2). Substitute the side lengths: 7^2 + 9^2 = 2 * (AD^2 + 4^2) ==> 49 + 81 = 2 * (AD^2 + 16) ==> 130 = 2 * (AD^2 + 16) ==> 65 = AD^2 + 16 ==> AD^2 = 65 - 16 = 49. Taking the square root: AD = sqrt(49) = 7 cm. Thus, the length of the median AD is 7 cm.

Solution 3: We apply Stewart's Theorem to triangle ABC with cevian AD of length d. The side lengths are c = AB = 5 cm, b = AC = 7 cm, and total base a = BC = 6 cm. The cevian divides the base into m = BD = 4 cm and n = DC = 2 cm (notice m + n = 4 + 2 = 6 = a). Stewart's Theorem states: b^2 * m + c^2 * n = a * (d^2 + m * n). Substitute the numerical values: 7^2 * (4) + 5^2 * (2) = 6 * (d^2 + 4 * 2) ==> 49 * 4 + 25 * 2 = 6 * (d^2 + 8) ==> 196 + 50 = 6 * (d^2 + 8) ==> 246 = 6 * (d^2 + 8). Divide by 6: 41 = d^2 + 8 ==> d^2 = 41 - 8 = 33. Taking the square root: d = sqrt(33) cm. Thus, the exact length of the cevian AD is sqrt(33) cm.