Stewart's theorem states that in the triangle shown below,
$$ b^2 m + c^2 n = a (d^2 + mn). $$
Is there any good way to prove this without using any trigonometry? Every proof I can find uses the Law of Cosines.
Geometry – Proving Stewart’s Theorem Without Trigonometry
geometrytriangles
Related Solutions
[As requested, posting a comment as a full answer.]
Taken from my answer to the question "What is the most elegant proof of the Pythagorean Theorem?" ...
Cut-the-Knot's SSS proof page has a number of solutions, including Euclid's. As the author indicates, however, only Hadamard's proof "goes through without a hitch", with the important aside: "assuming of course that isosceles triangles have been fairly treated previously".
I'll give a full development of Hadamard's argument, including the necessary bits needed about isosceles triangles. I include a couple of "obvious" sub-proofs just to make clear which axioms are in play.
Preliminaries:
- SAS triangle congruence is an axiom.
- (1) implies one direction of the Isosceles Triangle Theorem, namely: If two sides of a triangle are congruent, then the angles opposite those sides are congruent. $[\star]$
- (2) implies that A point equidistant from distinct points $P$ and $Q$ lies on the perpendicular bisector of the $\overline{PQ}$. $[\star\star]$
Now, we'll cut to the part of Hadamard's argument where we have $\triangle ABC$ and $\triangle A^\prime BC$ with corresponding edges congruent, constructed with $A$ and $A^\prime$ on the same side of $\overleftrightarrow{BC}$. (That last part is key.) We need only show that $A$ and $A^\prime$ coincide to prove SSS. Hadamard makes this argument by contradiction ...
- Assume $A \neq A^\prime$.
- $B$ is equidistant from $A$ and $A^\prime$, and therefore, $B$ lies on the perpendicular bisector of $\overline{AA^\prime}$ by (3) above. The same is true of $C$.
- Therefore, $\overleftrightarrow{BC}$ is the perpendicular bisector of $\overline{AA^\prime}$.
- Therefore, $\overleftrightarrow{BC}$ separates $A$ and $A^\prime$.
- This contradicts the fact that we constructed $A$ and $A^\prime$ to be on the same side of $\overleftrightarrow{BC^\prime}$. $[\star\star\star]$
- Therefore, the assumption that $A \neq A^\prime$ must be invalid, so that triangles $\triangle ABC$ and $\triangle A^\prime BC$ coincide; this gives us SSS. $\square$
$[\star]$ Proof: Isosceles $\triangle ABC$ with base $\overline{BC}$ is congruent to $\triangle ACB$ by SAS, and those opposite angles are corresponding angles of the congruent triangles.
$[\star\star]$ The midpoint, $M$, of $\overline{PQ}$ is certainly on the perpendicular bisector. A point $R \neq M$ equidistant from $P$ and $Q$ creates isosceles $\triangle RPQ$ with congruent angles at $P$ and $Q$ by (2). Thus, $\triangle RPM \cong \triangle RQM$ by SAS. Corresponding angles $\angle RMP$ and $\angle RMQ$ must be congruent; as supplements, they must be right. $\overleftrightarrow{MR}$, then, is the perpendicular bisector of $\overline{PQ}$. (I'll note that an easier proof could assert that, from the get-go, $\triangle RPM \cong RQM$ by SSS ... but we can't use that argument in a proof of SSS itself.)
$[\star\star\star]$ The contradiction here is one of Euclid's oversights later made explicit by Hilbert as the Plane Separation Axiom. The PSA has slightly-different formulations in different treatments of geometric foundations, but effectively it's the "Because I said so!" assertion that lines break planes into two disjoint "sides" and gives us our contradiction.
Best Answer
You can do it using the Pythagorean theorem. I'll treat the case in which $\angle (a,b)$ and $\angle(a,c)$ are both acute and $m \gt n$ as in your figure.
Draw the height $h$ of the triangle $abc$ and denote the resulting middle segment on $a$ by $x$. Using the Pythagorean theorem one sees
$$\begin{align*} b^2 &= h^2 + (n+x)^2 \\ c^2 &= h^2 + (m-x)^2 \\ d^2 &= h^2 + x^2 \end{align*}$$
Therefore $$\begin{align*} b^2 m &= h^2m + n^2m + 2mnx+x^2m \\ c^2 n &= h^2n + m^2n - 2mnx + x^2n \\ b^2 m + c^2n &= (n+m)(h^2 + mn + x^2) = a(d^2 + mn) \end{align*}$$ as we wanted.
If $\angle(a,c)$ is obtuse, the same idea works: Write $$b^2 = (m+n+x)^2 + h^2, \qquad c^2 = x^2 + h^2, \qquad d^2 = (m+x)^2 + h^2$$ and calculate similarly.
Symmetry and the consideration of some degenerate cases easily lead to a complete proof of Stewart's theorem from what I've written here.
It would be very nice to have a good pictorial proof that makes the identity similarly obvious as Pythagoras, but so far I couldn't come up with a nice figure that would achieve this.
Added: Matthew Stewart (1717–1785) published this theorem as Proposition II on page 2 in his 1746 book Some general theorems of considerable use in the higher parts of mathematics (arxive.org). Unfortunately, Google's scans don't seem to include the figures, but they are easy to reconstruct from the description in the text.
Update:(by @brainjam) I've taken the liberty of adding a partial scan of the figures from another Google scan -
One case of Stewart's Proposition II reads in the above notation:
$$b^2 + c^2 \frac{n}{m} = an + d^2 \frac{a}{m}$$
and it has the following nice corollary:
$$a^2 + b^2 + c^2 = 2d^2 + m^2 + n^2.$$
Here's the title page and the relevant passage (not using trigonometry either). Since Proposition I is used in the proof, I'm including the entire beginning of the text: