Question

proof of Pythagoras theorem​

Answer 1

Proof of pythagoras theorem is
AC^2=AB^2 +BC^2

Answer 2

»ɢɪᴠᴇɴ :-

• A right angled triangle ∆ ABC .
• The triangle is right angled at A .

»ᴛᴏ ᴘʀᴏᴠᴇ :-

• BC² = AB² + AC² that is the sum of square of base and perpendicular is equal to the square of hypontenuse .

»ᴄᴏɴsᴛʀᴜᴄᴛɪᴏɴ :-

• From vertex C drop a perpendicular on side BC of ∆ ABC . Mark it as point D .

»ᴘʀᴏᴏғ :-

$\underline{\textsf{\textbf{\red{\purple{ \dag }\:\:Figure:-}}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(15,8)\put(0,0){\line(1,0){5}}\put(0,5){\line(0,-1){5}}\put(5,0){\line(-1,1){5}}\put(2.5,2.6){\line(-1,-1){2.5}}\put(0.36,0){\line(0,1){0.36}}\put(0.36,0.36){\line(-1,0){0.36}}\put(0,5.3){ \bf C }\put(0,-0.3){ \bf A }\put(5,-0.3){ \bf B }\put(2.7,2.7){ \bf D }\put(2.8,2.2){\line(-1,-1){0.36}}\put(2.45,1.85){\line(-1,1){0.36}}\end{picture}$

We will prove this Theorem using similarity of triangles . Here we will try to Prove ∆ABC $\sim$ ∆ABD .

$\underline{\textsf{\textbf{\purple{\red{ \dag }\:\:In\: \triangle \:ABC\:\&\: \triangle \:DBA:-}}}}$

$\qquad\qquad\red{\tt \leadsto \angle ADB=\angle BAC\:\purple{(each\:equal\:to\:90^{\circ})}}$

$\qquad\qquad\red{\tt \leadsto \angle ABC=\angle DBA\:\purple{(common)}}$

$\sf \therefore \green{By\:AA\: similarity\: criterion}\:\\\sf \pink{\triangle ABC\:\sim\:\triangle DBA}$

$\underline{\sf\red{\mapsto} So\:now\:we\:can\:say\: that:-}$

$\tt:\implies \dfrac{BD}{AB}=\dfrac{AB}{BC}$

$\tt:\implies AB\times AB= \lgroup BC \times BD\rgroup$

$\boxed{\bf\red{\longmapsto}\:AB^2=BC\times BD }$

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$\sf\orange{ Similarly \:we \:can \:prove\: that\: \triangle ABC\: \sim \triangle ADC.}$

$\tt:\implies \dfrac{CD}{AC}=\dfrac{AC}{BC}$

$\tt:\implies AC\times AC= \lgroup BC \times CD \rgroup$

$\boxed{\bf\red{\longmapsto}\:AB^2=BC\times CD }$

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$\underline{\textsf{\textbf{\purple{\red{ \dag }\:\: Adding\:both \:the\:\: equations:-}}}}$

$\tt:\implies AB^2+ AC^2=(BC\times BD)+(BC\times CD)$

$\tt:\implies AB^2+AC^2=BC(CD+BD)$

$\tt:\implies AB^2+AC^2=BC\times BC$

$\underset{\pink{\bf Pythagoras\: Theorem}}{\purple{\underbrace{\underline{\boxed{\red{\tt\longmapsto\:\:AB^2\:+\:AC^2\:=\:BC^2}}}}}}$