Question

prove that: in a right angled triangle the square of the hypotenuse is equal to the sum of square of remaining two sides.​

Answer 1

Answer:

Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the remaining two sides.

Proof:

Given: In ΔABC,  m∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

In △ABC,

m∠ABC=90°                                                                  (Given)

seg BD is perpendicular to hypotenuse AC              (Construction)

Therefore, △ADB∼△ABC∼△BDC                             (Similarity of right-angled triangle)

△ABC∼△ADB

        (AB/AC)=(AD/AB)                                                (congruent sides of similar triangles)

 AB2=AD×AC                                                          (1)

△BDC∼△ABC

CD/BC=BC/AC                                                        (congruent sides of similar triangles)

BC2=CD×AC                                                            (2)

Adding the equations (1) and (2),

AB2+BC2=AD×AC+CD×AC

AB2+BC2=AC(AD+CD)  

Since, AD + CD = AC

Therefore, AC2=AB2+BC2

Hence Proved.  

There also exists a Converse of the Pythagoras theorem that states, “If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle”.

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Answer 2

Answer:

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