Understanding the Pythagoras Theorem. Click on the link to Watch the VIDEO explanation:
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Pythagoras Theorem
In a right - angled triangle, the square on the hypotenuse is equal to the sum of the squares on the sides containing the right angle.
Given ABC is a triangle with angle BAC is equal to 90 degrees
To prove that BC square is equal to AB square plus AC square
Construction On AB, BC and CA as sides describes squares ABFG, BCDE and ACKL respectively.
Draw AMN parallel to BE meeting BC at M and DE at N.
Join FC and AE.
The proof of the theorem are as follows
It is given that angle BAC is equal to 90 degrees
Angle BAG is equal to 90 degrees since it is one of the angles of square BAGF.
Therefore angle BAC plus angle BAG is equal to 180 degrees
Therefore CAG is a straight line
Angle ABF is equal to angle CBE since they are angles of squares and enhance each angle is equal to 90 degrees
Adding angle ABF to both sides
angle ABF plus angle ABC is equal to angle CBE plus angle ABC
This means angle CBF is equal to angle ABE
In triangles CBF and ABE, BF is equal to AB since they are the sides of the square BAFG
Recall the angle CBF is equal to angle ABE that we have proved earlier in the proof
BC is equal to BS since they are the sides of the square BCDE
Therefore, triangle CBF is congruent to triangle ABE by SAS congruency
Therefore, triangle CBF is equal to triangle ABE
But triangle CBF is equal to half square ABFG Since they have the same base BF and they are between the same parallels BF, CG.
Triangle ABE is equal to half rectangle BENM Since they have the same base BE and are between the same parallels BE and AN.
Therefore, BENM is equal to square ABFG Since triangle CBF equal to triangle ABE
Similarly, rectangle CDNM is equal to square ACKL
Therefore BENM plus CDNM is equal to square ABFG plus square AKCL
This means square BCDE is equal to square ABFG plus ACKL
Therefore, BC square is equal to AB square plus AC square
Thus, we have proved the theorem.
Watch Video
Pythagoras Theorem
In a right - angled triangle, the square on the hypotenuse is equal to the sum of the squares on the sides containing the right angle.
Given ABC is a triangle with angle BAC is equal to 90 degrees
To prove that BC square is equal to AB square plus AC square
Construction On AB, BC and CA as sides describes squares ABFG, BCDE and ACKL respectively.
Draw AMN parallel to BE meeting BC at M and DE at N.
Join FC and AE.
The proof of the theorem are as follows
It is given that angle BAC is equal to 90 degrees
Angle BAG is equal to 90 degrees since it is one of the angles of square BAGF.
Therefore angle BAC plus angle BAG is equal to 180 degrees
Therefore CAG is a straight line
Angle ABF is equal to angle CBE since they are angles of squares and enhance each angle is equal to 90 degrees
Adding angle ABF to both sides
angle ABF plus angle ABC is equal to angle CBE plus angle ABC
This means angle CBF is equal to angle ABE
In triangles CBF and ABE, BF is equal to AB since they are the sides of the square BAFG
Recall the angle CBF is equal to angle ABE that we have proved earlier in the proof
BC is equal to BS since they are the sides of the square BCDE
Therefore, triangle CBF is congruent to triangle ABE by SAS congruency
Therefore, triangle CBF is equal to triangle ABE
But triangle CBF is equal to half square ABFG Since they have the same base BF and they are between the same parallels BF, CG.
Triangle ABE is equal to half rectangle BENM Since they have the same base BE and are between the same parallels BE and AN.
Therefore, BENM is equal to square ABFG Since triangle CBF equal to triangle ABE
Similarly, rectangle CDNM is equal to square ACKL
Therefore BENM plus CDNM is equal to square ABFG plus square AKCL
This means square BCDE is equal to square ABFG plus ACKL
Therefore, BC square is equal to AB square plus AC square
Thus, we have proved the theorem.
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