Question

prove the "Pythagoras Theorem"​

Answer 1

Answer:

Pythagoras theorem states that “ In a right-angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”. The sides of the right-angled triangle are called base, perpendicular and hypotenuse . Proof: ... ∠BAD = ∠BAC i.e. ∠A is common in both triangles.

Answer 2

Answer:

Proof:

Proof:We know, △ADB ~ △ABC

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABAC

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABC

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × AC

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × ACAB2 + BC2 = AC (AD + CD)

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × ACAB2 + BC2 = AC (AD + CD)Since, AD + CD = AC

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × ACAB2 + BC2 = AC (AD + CD)Since, AD + CD = ACTherefore, AC2 = AB2 + BC2

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × ACAB2 + BC2 = AC (AD + CD)Since, AD + CD = ACTherefore, AC2 = AB2 + BC2Hence, the Pythagorean theorem is proved.

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABACOr, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (corresponding sides of similar triangles)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × ACAB2 + BC2 = AC (AD + CD)Since, AD + CD = ACTherefore, AC2 = AB2 + BC2Hence, the Pythagorean theorem is proved.Note: Pythagorean theorem is only applicable to Right-Angled triangle.