Question

state and prove the coverse of pythagoras theorem

Answer 1

Answer:

Step-by-step explanation:

Given :  

A triangle ABC such that AB² + BC²  =  AC²

To Prove :

ΔABC is right-angled at B.

Construction :  

Construct a right-angled triangle PQR, right-angled at Q such that PQ  =  AB and QR  =  BC.  

solution;

Step 1 :  

In ΔPQR, ∠Q = 90°.  

Using Pythagorean theorem in ΔPQR, we have  

PQ² + QR²  =  PR² ----- (1)

Step 2 :  

In ΔABC (given), we have  

AB² + BC²  =  AC² ----- (2)

Step 3 :  

By construction,  PQ  =  AB and QR  =  BC.

So, from (1) and (2), we have

PR²  =  AC²

Get rid of the square from both sides.

PR  =  AC

Step 4 :  

Therefore, by SSS congruence criterion, we get

ΔABC  ≅  ΔPQR

which gives

∠B  =  ∠Q  

Step 5 :  

But, we have ∠Q = 90° by construction.  

Therefore ∠B = 90°.

Hence, ΔABC is a right triangle, right angled at B.  

Thus, the theorem is proved.

Answer 2

Step-by-step explanation:

Statement:

In a Triangle the square of longer side is equal to the sum of squares of the other two sides, then the triangle is a right angled triangle.

Given -

A Triangle ABC such that

BC² = AB² + AC²

To Prove -

Angle A = 90°

Construction -

Draw a ∆DEF such that AB = DE and AC = DF and Angle D = 90°

Proof -

In ∆ABC,

BC² = AB² + AC² - Given

In ∆ DEF

EF² = DE² + DF²

Therefore,

EF² = AB² + AC²

(Since AB = DE, AC = DF)

Therefore,

BC² = EF² ie - BC = EF

Now, In ∆ABC and ∆DEF

AB = DE - By Construction

AC = DF - By Construction

BC = EF

Therefore

∆ABC ≅ ∆DEF by SSS test.

Thus,

Angle A = Angle D - CPCT

But, Angle D = 90° ( As per construction)

Therefore

Angle A = 90°

Hence Proved! .....................