If triangle ABC is a right angled triangle at B,then AB² + BC² = AC²

Given : A right ΔABC right angled at B

To prove : AC² = AB² + BC²

Construction : Draw AD ⊥ AC

Proof : ΔABD and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC  (common)

∴ ΔADB ∼ ΔABC  (by AA similarly criterion)

$\Rightarrow \frac{\mathrm{AD}}{\mathrm{AB}} = \frac{\mathrm{AB}}{\mathrm{AC}} \left[\mathrm{Corresponding} \mathrm{sides} \mathrm{of} \mathrm{similar} ∆\text{'}\mathrm{s} \mathrm{are} \mathrm{proportional}\right]$

⇒ AD × AC = AB²    ...... (1)

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA  (common)

∴ ΔBDC ∼ ΔABC  (by AA similarly criterion)
 

$\Rightarrow \frac{\mathrm{BC}}{\mathrm{AC}} = \frac{\mathrm{CD}}{\mathrm{BC}} \left[\mathrm{Corresponding} \mathrm{sides} \mathrm{of} \mathrm{similar} ∆\text{'}\mathrm{s} \mathrm{are} \mathrm{proportional}\right]$   

⇒ CD × AC = BC²    ........ (2)
 

Adding (1) and (2) we get

AB² + BC² = AD × AC + CD × AC

= AC (AD + CD)

= AC × AC = AC²

∴ AC² = AB² + BC²