Question

Heya mates! Gimme proof and statement of Converse of Pythagoras theorem! ^^"

Answer 1

$\huge\underline\mathfrak\red{Statement}$

In a triangle, if the square of one side is equal to the sum of square of other two sides then prove that the triangle is right angled triangle.

________________________

$\huge\underline\mathfrak\red{Solution}$

Given : AC² = AB² + BC²

To prove : ABC is a right angled triangle.

Construction : Draw a right angled triangle PQR such that, angle Q = 90°, AB = PQ, BC = QR.

Proof : In triangle PQR,

Angle Q = 90° ( by construction )

Also,

PR² = PQ² + QR² ( By using Pythagoras theorem )...(1)

But,

AC² = AB² + BC² ( Given )

Also, AB = PQ and BC = QR ( by construction )

Therefore,

AC² = PQ²+ QR²....(2)

From eq (1) and (2),

PR² = AC²

So, PR = AC

Now,

In ∆ABC and ∆PQR,

AB = PQ ( By construction )

BC = QR ( By construction )

AC = PR ( Proved above )

Hence,

∆ABC is congruent to ∆PQR by SSS criteria.

Therefore, Angle B = Angle Q ( By CPCT )

But,

Angle Q = 90° ( By construction )

Therefore,

Angle B = 90°

Thus, ABC is a right angled triangle with Angle B = 90°

____________________

Hence proved!

Answer 2

SOLUTION

STATEMENT:-

In a ∆, if the sum of the squares of 2 sides is equal to square of third side, then angle opposite to third side is right angle.

GIVEN:-

ABC is a ∆ in which AC^2= AB^2+ BC^2

To Prove:-

Angle B= 90°

Construction:-

Draw a ∆PQR right angled at Q, such that QR= BC & PQ= AB.

Proof:-

In ∆ABC,

${AC}^{2} = {AB}^{2} + {BC}^{2} \: \: \: (given)...........(1) \\ now \: PQ = AB \: \: \: \: [by \: construction] \\ QR = BC \\ \\ from \: (1) \\ {AC}^{2} = {PQ}^{2} + {QR}^{2} ............(2) \\ Now, \: from \: \triangle \: PQR \\ {PR}^{2} = {PQ}^{2} + {QR}^{2} \: \: \: (pythagoras \: theorem) \\ from \: (2) ,\: we \: get \\ {AC}^{2} = {PR}^{2} = > AC = PR \\ \\ Now, \: in \: \triangle \: ABC \: and \: \triangle \: PQR \\ = > AB= PQ \: \: [by \: construction] \\ = > BC = QR \: \: \: \: [by \: construction]\\ = > AC= PR \: \: [proved \: above] \\ \\ = > ABC ≅ PQR \: [SSS] \\ = > \angle \: B= \angle \: Q \: \: \: [cpct] \\ = > \angle \: B= 90 \degree$

Hope it helps ☺️