Question

# State and prove Pythagoras theorem.

Solution

## Given triangle ABC is a right angle triangle at BDraw BD perpendicular to ACIn $$\Delta ABC$$ 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. Mathematics

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