Question

In a triangle if, the square of one side is equal to the sum of the squares of the other two sides,then the angle opposite to the first side is right angle.prove

Answer 1

Hey there !!


→ Given :- A ∆ABC in which AC² = AB² + BC².

→ To prove :- $\angle B$ = 90° .

→ Construction :- Draw a ∆DEF such that DE = AB, EF = BC and $\angle E$ = 90°.

→ Proof :-)


▶ In ∆DEF , we have $\angle E$ = 90°.

So, by Pythagoras' theorem , we have

=> DF² = DE² + EF².

=> DF² = AB² + BC². ........(1).
[ => DE = AB and EF = BC. ]


But,
AC² = AB² + BC². ......... (2). [Given]


➡ From equation (1) and (2), we get

=> AC² = DF².
[ By taking under root both side, we get ]

=> AC = DF.


▶Now, in ∆ABC and ∆DEF, we have

=> AB = DE. [By construction]

=> BC = EF. [By construction]

=> AC = DF. [ By showed above]

→ By SSS congruency criteria,

$\bf{ \therefore \triangle ABC \cong \triangle DEF . }$

→ Hence, $\huge \boxed{ \boxed{ \bf \angle B = \angle E = 90 \degree }}$


✔✔ Hence, it is proved ✅✅.

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$\huge \boxed{ \boxed{ \boxed{ \mathbb{THANKS}}}}$



$\huge \bf{ \# \mathbb{B}e \mathbb{B}rainly.}$

Answer 2

HEY MATE HERE IS YOUR ANSWER

TO PROVE = the opposite angle to the first side is right angle

Proof :

In ⛛ DEF, we have /_90°

so, by Pythagoras theorem

$= > {df}^{2} = {de}^{2} + {ef}^{2}$
$= > \: {df}^{2} = {ab}^{2} + {bc}^{2} - - - - (1)$
$= > de = ab \: and \: eb = bc$

But

AC^2 =AB^2=BC^2 - - - - - - (2)(GIVEN)

FROM EQUATION (1) AND (2) WE GET

=> AC^2 = DF^2

=> AC = DF

Now in ⛛ ABC AND DEF

WE HAVE

AB = DE (GIVEN}
BC = EF (GIVEN)
AC = DF { PROVED ABOVE}

⛛ ABC ~ ⛛ DEF ( BY SSS Congruence)

HENCE, /_B=/_E = 90°

PROVED

HOPE IT WILL HELP YOU