In this chapter, we will attempt to prove Pythagoras theorem formula using basic geometry and algebra concepts.

The Pythagoras formula proof is frequently asked in math exams so make sure you practice the below method.

Let’s first revise the Pythagoras formula.

## What is Pythagoras theorem formula ?

It states that for an right angled triangle, the **square of hypotenuse is equal to sum of square of other two sides**.

\mathtt{hypotenuse^{2} =\ ( side\ 1) \ ^{2} +\ ( side\ 2)^{2}}\\\ \\ \mathtt{AC^{2} =\ AB^{2} +\ BC^{2}}

## Proof of Pythagoras theorem

Consider the triangle ABC, right angled at B.

Now **draw perpendicular line BO from vertex B to line AC**.

Here, ∠BOC = 90 and ∠BOA = 90 degree

Now consider **triangle AOB and ABC**.

∠A = ∠A { common angle }

∠AOB = ∠ABC = 90 degree.

By AA similarity condition, **both the triangles are similar**.

Hence, \mathtt{\triangle AOB\ \sim \ \triangle ABC}

When the triangles are similar then their **sides are in proportion**.

\mathtt{\frac{OC}{BC} =\frac{BC}{AC}}\\\ \\ \mathtt{BC^{2} \ =\ OC\ \times \ AC} – – – eq(1)

Similarly taking** triangle COB and ABC**

∠C = ∠C { common angle }

∠COB = ∠ABC = 90 degree.

By AA similarity condition, **both the triangles are similar.**

Hence, \mathtt{\triangle COB\ \sim \ \triangle ABC}

When the triangles are similar then their **sides are in proportion**.

\mathtt{\frac{OC}{AB} =\frac{AB}{AC}}\\\ \\ \mathtt{AB^{2} \ =\ OA\ \times \ AC} – – -eq(2)

**Adding both eq (1) and eq (2)**

\mathtt{AB^{2} \ +\ BC^{2} =\ OA\ \times \ AC\ +\ OC\ \times \ AC}\\\ \\ \mathtt{AB^{2} \ +\ BC^{2} \ =\ AC\ ( OA\ +\ OC)}\\\ \\ \mathtt{AB^{2} \ +\ BC^{2} \ =\ AC\ ( AC)}\\\ \\ \mathtt{AB^{2} \ +\ BC^{2} \ =\ AC^{2}}

**Hence Proved.****Note:**

This is algebra method of Pythagoras formula proof.

You can also prove the formula with help of basic geometry.

### Proving Pythagoras theorem using geometry

Consider a **square ABCD of side b + c cm** inside which a **square PQRS of side a cm** is drawn as shown below.

We can see that; **Area of square ABCD = Area of square PQRS + area of 4 right triangles** ( i.e. ASP + SDR + RCQ + PBQ ) – eq (1)

**Area of square ABCD** = \mathtt{( b+c)^{2}} – eq (2)**Area of square PQRS** = \mathtt{(a)^{2}} – eq (3)**Area of 4 right triangles**;

\mathtt{\Longrightarrow \ 4\ \times \frac{1}{2} \times b\times c}\\\ \\ \mathtt{\Longrightarrow \ 2bc} – eq (4)

Put eq (2), (3) & (4) in equation (1)

\mathtt{( b+c)^{2} \ =\ a^{2} +2bc}\\\ \\ \mathtt{b^{2} +c^{2} +2bc\ =\ a^{2} +2bc}\\\ \\ \mathtt{b^{2} +c^{2} +\cancel{2bc} =a^{2} +\cancel{2bc}}\\\ \\ \mathtt{a^{2} =b^{2} +c^{2}}

**Hence we proved the Pythagoras theorem.**

You can apply this formula in above triangles like QCR, PBQ etc.