Question

prove Pythagoras theorem​

Answer 1

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Answer 2

Answer:

Proof:

Proof:We know, △ADB ~ △ABC

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABAC (corresponding sides of similar triangles)

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABAC (corresponding sides of similar triangles)Or, AB2 = AD × AC ……………………………..……..(1)

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

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

Proof:We know, △ADB ~ △ABCTherefore, ADAB=ABAC (corresponding sides of similar triangles)Or, 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=ABAC (corresponding sides of similar triangles)Or, 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=ABAC (corresponding sides of similar triangles)Or, 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=ABAC (corresponding sides of similar triangles)Or, 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=ABAC (corresponding sides of similar triangles)Or, 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=ABAC (corresponding sides of similar triangles)Or, 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=ABAC (corresponding sides of similar triangles)Or, 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=ABAC (corresponding sides of similar triangles)Or, 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.