Question

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Prove the theorem

Converse of Pythagoras theorem
Prove it

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Answer 1

Answer:

Your answer is here mate :-

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Given -AC^2=AB^2+BC^2

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To prove -<B =90°

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Cons- Draw a rt triangle PQR similar to ∆ABC such that AB=PQ,BC=QR

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Proof-

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IN PQR:-

PR^2=PQ^2+QR^2 [by phytgagorus th^m]

But PQ=AB,BC=QR[given]

So,

PR^2=AB^2+BC^2. -(eq1)

But ,AC^2=AB^2+ BC^2[given]. -(eq2)

From e/q 1 and 2

PR =AC

IN ∆ABC and ∆PQR

AB =PQ(given)

BC=QR(given)

AC=PR(proved above)

∆ABC is congruent to ∆PQR(by SSS)

So <ABC=<PQR

= <PQR=90°{given}

So,

= <ABC =90°

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hope it helps uh

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Answer 2

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$\huge{\red{\mathfrak{ANSWER:-}}}$

Converse of Pythagoras theorem

Step-by-step explanation:

So

Statement :- 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 side is a right angle

Given :- A triangle ABC such that

${ac}^{2} = {ab}^{2} + {bc}^{2}$

Construction :-

Draw a triangle DEF such that

DE= AB, EF=BC and angle E= 90 degree

Proof:-

So in order to prove that angle B = 90 degree, for that we to prove that

Triangle ABC similar triangle DEF

Since,

We have,

DEF is a right angled triangle

Therefore,

By using Pythagoras theorem

${df}^{2} = {de}^{2} + {ef }^{2} \\ {df}^{2} = {ab}^{2} + {bc}^{2} \\ {df}^{2} = {ac}^{2} \\ df = ac \: \: \: \: \: \: \: \: \: ...........equation \: 1$

And

we have

In triangle ABC and triangle DEF

$ab = de \: \\ bc = ef \: \: \: \: \: (by \:contruction)\\ ac = df \: \: (from \: equation \: 1)$

Therefore,

$triangle \: abc \: similiar \: triangle \: def \: \: \: \: \: \: (\: by \: sss \: criteria)$

Hence

Angle B =Angle E = 90 degree

Hence,

Triangle ABC is a triangle right-angled at B

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