Stewart’s Theorem

What is Stewart’s Theorem?

  • Stewart's theorem in Geometry yields a relation between the cervain length and the side lengths of a triangle. 

  • Stewart's Theorem can be proved using the law of cosines as well as by using the famous Pythagorean Theorem. 

  • The theorem was proposed in honor of the Scottish mathematician Matthew Stewart in 1746.

Stewart’s Theorem Proof:

Here’s the Stewart’s Theorem proof,

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In △ABC, point D is a point on BC where AB=c, AC=b, BD=u, DC=v, AD=t. 

Stewart's theorem states that in this triangle, the following equation given below holds:

t² = [ (b²u+c²v)/(u+v) ]- uv


In Stewart’s Theorem Geometry, the Stewart’s theorem can be proved by drawing the perpendicular from the vertex of the triangle up to the base and by using the Pythagoras Theorem for writing the distances b, d, c, in terms of altitude. The right and left-hand sides of the equation reduce algebraically to form the same kind of expression.

Before moving on to the proof by Pythagoras Theorem, let’s know what Pythagoras Theorem is!

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The little box in the corner of the triangle denotes the right angle which is equal to 90 degrees.

● The side opposite to the right angle is the longest side of the triangle which is known as the hypotenuse (H).

● The side that is opposite to the angle is known as the opposite (O).

● And the side which lies next to the angle is known as the Adjacent (A)

According to Pythagoras theorem, 

In a right-angle triangle,

(Opposite)2+(Adjacent)2 = (Hypotenuse)2


Proof by Pythagoras Theorem-

Let us assume ∠B and ∠C are both acute angles and where u<v in the figure given above. Then we have,

 t2 =h2+ x2

b2 =h2+ (v-x)2 = b2u = h2 u + uv2 – 2uvx+ ux2 

c2 =h2+ (u+x)2 = c2u = h2 v+ u2 v+ 2uvx+ vx2 

=b2 u+ c2 u =h2u+ h2v+ uv2 + u2v -2uvx +2uvx + vx2+ ux2

= (u +v) (h2+uv+ x2)

= (u+v)( t2+uv)

=a(t2+uv)

Stewart’s Theorem Statement in Stewart theorem Geometry

 If a, b, c are the lengths of the triangle ABC. Let d be the length of the cevian of the side of the length a. Suppose the cevian d divides the side ‘a’ into 2 segments of the length m and n, where m is equal to the adjacent to the side c and whereas n is equal to the adjacent to the side b, then prove that-

b2m+c2 n=a(d2+mn)

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Proof:

The theorem can be completed by using the law of cosines.

Let us consider θ to be the angle between the side m and side ‘d’ and Let θ′ be equal to the angle between the side n and side d. This angle θ′ is usually the addition of θ and cos θ′ = (− cos θ).

The Law of the cosines for the angles θ′ and θ states that-

c2=m2+d2−2dmcos θ

b2=n2+d2−2dncos θ ‘

=n2+d2+2dncos θ

Now, we need to multiply the first equation by n, and the second equation needs to be multiplied by m, and later add the two equations to remove cos θ, 

Now we obtain-

b2 m+c2 n

=nm2+n2m+ (m+ n) d2

= (m+n) (mn+d2)

=a (mn+d2) =  b2 m+c2 n

Therefore, this is the required proof.

Now you might think what a cevian is?

In geometry, a cevian can be defined as any line segment in a triangle with one endpoint on a vertex of the triangle and the other endpoint on the opposite side of the triangle. Special cases of cevians include medians, altitudes, and angle bisectors .

Trick to remember the Stewart Theorem in Stewart theorem Geometry-

To remember this theorem, use this trick,

(man+dad) = (bmb+cnc)

Or a man (product of m,a,n) and his dad( product of a,d and d) put a bomb ( product of b, m and b) in the sink (cnc).

A man and his dad put a bomb in the sink.


Uses of Stewart’s Theorem-

In Stewart Theorem Geometry, Stewart's theorem defines the relationship between the lengths of sides of any given triangle as well as the length of the cevian of the triangle.

Problems to be solved:

Question 1) Prove Stewart’s Theorem using Pythagoras Theorem.

Answer) Proof by Pythagoras Theorem-

Let us assume ∠B and ∠C are both acute angles and where u<v in the figure given above. Then we have,

 t2 =h2+ x2

b2 =h2+ (v-x)2 = b2u = h2 u + uv2 – 2uvx+ ux2 

c2 =h2+ (u+x)2 = c2u = h2 v+ u2 v+ 2uvx+ vx2 

=b2 u+ c2 u =h2u+ h2v+ uv2 + u2v -2uvx +2uvx + vx2+ ux2

= (u +v) (h2+uv+ x2)

= (u+v)( t2+uv)

=a(t2+uv) 

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FAQ (Frequently Asked Questions)

Q1. How do you Memorize Theorems and what is mn Theorem?

Ans. If you want to Remember any Theorem then there are a Few Things you Need to Remember:

  1. Try out the theorem in a computable example. If it's a classification theorem, then you need to pick some object and follow the steps of the proof on your chosen object.

  2.  You need to build examples and counterexamples.

  3. Try to remove the hypotheses of the theorem.

The mn Theorem can be defined as -

Let D be the point on the side BC of a triangle namely ABC that divides the side BC in the required ratio m: n, then we have: (m + n) cot θ = m cot α – n cot ß. (m + n) cot θ = n cot B – m cot C.

Q2. What is Half of a Triangle Called and what is a Cevian?

Ans. A triangle which has sides which are equal in length is known as an equilateral triangle,  triangle which has two sides of equal length are known as isosceles, and a triangle which has all sides a different length are known as scalene triangles. A triangle can be simultaneously right and isosceles, in this case it is known as an isosceles right triangle.

 

In geometry, a cevian can be defined as any line segment in a triangle with one endpoint on a vertex of the triangle and the other endpoint on the opposite side of the triangle. Special cases of cevians include medians, altitudes, and angle bisectors.