proof of Pythagoras theorem
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Proof of pythagoras theorem isAC^2=AB^2 +BC^2
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[tex]\underline{\textsf{\textbf{\red{\purple{$\dag$}\:\:Figure:-}}}}[/tex]
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We will prove this Theorem using similarity of triangles . Here we will try to Prove ∆ABC [tex]\sim [/tex] ∆ABD .
[tex]\underline{\textsf{\textbf{\purple{\red{$\dag$}\:\:In\:$\triangle$ \:ABC\:\&\:$\triangle$\:DBA:-}}}}[/tex]
[tex]\qquad\qquad\red{\tt \leadsto \angle ADB=\angle BAC\:\purple{(each\:equal\:to\:90^{\circ})}}[/tex]
[tex]\qquad\qquad\red{\tt \leadsto \angle ABC=\angle DBA\:\purple{(common)}}[/tex]
[tex]\sf \therefore \green{By\:AA\: similarity\: criterion}\:\\\sf \pink{\triangle ABC\:\sim\:\triangle DBA} [/tex]
[tex]\underline{\sf\red{\mapsto} So\:now\:we\:can\:say\: that:-} [/tex]
[tex]\tt:\implies \dfrac{BD}{AB}=\dfrac{AB}{BC}[/tex]
[tex]\tt:\implies AB\times AB= \lgroup BC \times BD\rgroup[/tex]
[tex]\boxed{\bf\red{\longmapsto}\:AB^2=BC\times BD }[/tex]
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[tex]\sf\orange{ Similarly \:we \:can \:prove\: that\: \triangle ABC\: \sim \triangle ADC.}[/tex]
[tex]\tt:\implies \dfrac{CD}{AC}=\dfrac{AC}{BC}[/tex]
[tex]\tt:\implies AC\times AC= \lgroup BC \times CD \rgroup[/tex]
[tex]\boxed{\bf\red{\longmapsto}\:AB^2=BC\times CD }[/tex]
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[tex]\underline{\textsf{\textbf{\purple{\red{$\dag$}\:\: Adding\:both \:the\:\: equations:-}}}}[/tex]
[tex]\tt:\implies AB^2+ AC^2=(BC\times BD)+(BC\times CD)[/tex]
[tex]\tt:\implies AB^2+AC^2=BC(CD+BD)[/tex]
[tex]\tt:\implies AB^2+AC^2=BC\times BC[/tex]
[tex]\underset{\pink{\bf Pythagoras\: Theorem}}{\purple{\underbrace{\underline{\boxed{\red{\tt\longmapsto\:\:AB^2\:+\:AC^2\:=\:BC^2}}}}}} [/tex]