Prove pythagoras theorem?
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Pythagoras theorem is :
In a right angle triangle , sum of squares of sides is equal to the square of the hypotenuse.
We can solve that using similar triangles principles. This is the technique that Pythagoras used.
See the diagram enclosed.
ABC is a right angle triangle at B and BD is a perpendicular from B on to AC, the diagonal.
Triangles ABC and BDC are similar, as:
. angle DBC = 90 - angle C = angle A
. angle ABC = angle BDC = 90
Ratios of corresponding sides are equal.
AB / BD = BC / DC = AC / BC
So we get BC² = AC * DC --- (1)
Triangle ABC amd ADB are similar, as:
. angle DBA = 90 - (90 - C) = angle C
. angle ABC = 90 = angle ADB
so ratios of corresponding sides are equal
AB / AD = BC / BD = AC / AB
So we get AB² = AD * AC --- (2)
Add (1) and (2) to get:
AB² + BC² = AC * ( AD + DC) = AC * AC = AC²
So the theorem is proved.
In a right angle triangle , sum of squares of sides is equal to the square of the hypotenuse.
We can solve that using similar triangles principles. This is the technique that Pythagoras used.
See the diagram enclosed.
ABC is a right angle triangle at B and BD is a perpendicular from B on to AC, the diagonal.
Triangles ABC and BDC are similar, as:
. angle DBC = 90 - angle C = angle A
. angle ABC = angle BDC = 90
Ratios of corresponding sides are equal.
AB / BD = BC / DC = AC / BC
So we get BC² = AC * DC --- (1)
Triangle ABC amd ADB are similar, as:
. angle DBA = 90 - (90 - C) = angle C
. angle ABC = 90 = angle ADB
so ratios of corresponding sides are equal
AB / AD = BC / BD = AC / AB
So we get AB² = AD * AC --- (2)
Add (1) and (2) to get:
AB² + BC² = AC * ( AD + DC) = AC * AC = AC²
So the theorem is proved.
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Answer: Pythagoras' theorem: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. ... Similarly, construct two squares with sides measuring a units and b units along the sides and CA of the ΔABC.
Step-by-step explanation:That's in my knowledge
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