Stewart’s Theorem
Stewart’s Theorem, though outside the Hong Kong Exam syllabus, is one of the important
theorems in geometry. It is not too difficult to learn. See whether you can manage.
Given any triangle ABC. AD is any arbitrary line. The base of the triangle: a = m + n.
(a) Prove Stewart’s Theorem: b2 m + c 2 n = a(d2 + mn) by:
(i) Pythagoras Theorem
(ii) using trigonometry.
(b)
In the diagram on the right,
AD bisects ∠BAC .
AB = 8, AC = 6, BD = 4,
Find the length of AD .
(c) Prove that the sum of the square of the distances from the vertex of the right angle, in a right
angled triangle, to the trisection point of the hypotenuse, is equal to
hypotenuse:
5
d2 + e2 = 9 a2
5
9
the length of the
(a) (i)
Draw a perpendicular line AE from A to BC .
Let AE = h, ED = x, BE = m − x
Then:
b2 = h2 + (n + x)2 … . (1)
c 2 = h2 + (m − x)2 … . (2)
d2 = h2 + x 2
… . (3)
(1) − (3), b2 − d2 = n2 + 2nx … . (4)
(2) − (3), c 2 − d2 = m2 − 2mx … . (5)
(4) × m,
(5) × m,
b2 m − d2 m = n2 m + 2mnx … . (6)
c 2 n − d2 n = m2 n − 2mnx … . (7)
(6) + (7),
b2 m + c 2 n − d2 (m + n) = mn(m + n)
b2 m + c 2 n − ad2 = amn
b2 m + c 2 n = a(d2 + mn)
(ii) Let
∠ADB = α, ∠ADC = 180° − α
By Cosine Law,
cos α =
m2 +d2 −c2
2md
cos(180° − α) =
−cos α =
m2 +d2 −c2
2md
… . (1)
n2 +d2 −b2
n2 +d2 −b2
2nd
=−
2nd
… . (2)
n2 +d2 −b2
2nd
−m(n2 + d2 −b2 ) = n(m2 + d2 −c 2 )
Rearrange and putting a = m + n, we have
b2 m + c 2 n = a(d2 + mn)
Stewart’s Theorem can be re-written in the form:
man + dad = bmb + cnc . (Mnemonic : A man and his dad put a bomb in the sink.)
∠BAD = ∠CAD = θ
(b) Let
Since
Area of ∆ABD
Area of ∆ACD
1
×8dsin θ
2
1
×6dsin θ
2
4
=n
4
=n
∴n=3
(This is angle bisector theorem:
8
4
=n .)
6
By Stewart’s Theorem
b2 m + c 2 n = a(d2 + mn)
62 (4) + 82 (3) = 7(d2 + 4 × 3)
∴ d = 6 (Taking positive root.)
(c) Apply Stewart theorem to ∆ABC in two
ways, using AD and AE as the lines of
cutting the triangle:
2
1
2
1
c 2 (3 a) + b2 (3 a) = a [d2 + (3 a) (3 a)]
{
1
2
2
1
c 2 (3 a) + b2 (3 a) = a [e2 + (3 a) (3 a)]
2
1
2
1
c 2 (3) + b2 (3) = d2 + (3 a) (3 a)
{
1
2
2
1
c 2 (3) + b2 (3) = e2 + (3 a) (3 a)
4
Adding, we have c 2 + b2 = d2 + e2 + 9 a2
By Pythagoras Theorem,
4
a2 = d2 + e2 + a2
9
5
∴ d2 + e2 = 9 a2
Yue Kwok Choy
19/8/2016