how to prove the pythagoras theorem
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Answer:
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Pythagorean theorem, the well-known geometric theorem that the sum of the squares on the legs of a right triangle is equal to the square on the hypotenuse (the side opposite the right angle)—or, in familiar algebraic notation, a2 + b2 = c2
Step-by-step explanation:
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Pythagoras Theorem ProofGiven: A right-angled triangle ABC.To Prove- AC2 = AB2 + BC2Proof: First, we have to drop a perpendicular BD onto the side AC
Proof: First, we have to drop a perpendicular BD onto the side AC
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABC
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABC
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (Condition for similarity)
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (Condition for similarity)Or, BC2= CD × AC ……………………………………..(2)
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (Condition for similarity)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (Condition for similarity)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × AC
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (Condition for similarity)Or, BC2= CD × AC ……………………………………..(2)Adding the equations (1) and (2) we get,AB2 + BC2 = AD × AC + CD × ACAB2 + BC2 = AC (AD + CD)
Proof: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (Condition for similarity)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: First, we have to drop a perpendicular BD onto the side ACWe know, △ADB ~ △ABCTherefore, ADAB=ABAC (Condition for similarity)Or, AB2 = AD × AC ……………………………..……..(1)Also, △BDC ~△ABCTherefore, CDBC=BCAC (Condition for similarity)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
Hence, the Pythagorean thoerem is proved.
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