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Question

State and prove Pythagoras theorem.


Solution

Given triangle ABC is a right angle triangle at B

Draw BD perpendicular to AC

In $$\Delta ABC $$ and $$\Delta ADB$$

$$\angle ABC=\angle ADB $$

$$\angle A=\angle A$$

$$\therefore \Delta ABC\sim \Delta ADB$$

$$\therefore \dfrac{AB}{A D}=\dfrac{BC}{DB}=\dfrac{AC}{AB}$$

$$\Rightarrow AB^{2}=AC\times AD$$.............................................(1)

In$$\Delta ABC$$ and $$\Delta BDC$$

$$\angle ABC=\angle BDC $$

$$\angle C=\angle C$$

$$\therefore \Delta ABC\sim \Delta BDC$$

$$\therefore \dfrac{AB}{BD}=\dfrac{BC}{DC}=\dfrac{AC}{BC}$$

$$\Rightarrow BC^{2}=AC\times DC$$

Add (1) and (2) we get

$$AB^{2}+BC^{2}=AC\times AD+AC\times DC$$

$$=AC(AD+DC)=AC\times AC$$

$$=AC^{2}$$ 

Hence proved.
 


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Mathematics

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