[Boutique] Pythagorean theorem proof

Part 1: Proof of Pythagorean Theorem

Make two congruent right-angled triangles, set their two right-angled sides to a, b, and the bevel length to c. Then make a square with side length c. Put them together into a polygon as shown.

Square FCJI and AEIG are made with CF and AE as side lengths, respectively.

∵EF=DF-DE=ba, EI=b,

∴FI=a,

∴G, I, J are on the same line,

∵CJ=CF=a, CB=CD=c,

∠CJB = ∠CFD = 90o,

∴RtΔCJB ≌ RtΔCFD,

Similarly, RtΔABG ≌ RtΔADE,

∴RtΔCJB ≌ RtΔCFD ≌ RtΔABG ≌ RtΔADE

∴∠ABG = ∠BCJ,

∵∠BCJ +∠CBJ= 90o,

∴∠ABG +∠CBJ= 90o,

∵∠ABC= 90o,

∴G, B, I, J are on the same line,

Article 2: Proof of Pythagorean Theorem

Solution: In the ruled line, the area of ​​a small square with two right-angled sides as the side length is equal to the square area with the oblique side as the side length.

The content of the Pythagorean theorem: the square of the two right-angled sides of a right-angled triangle a, b is equal to the square of the oblique-edge c.

a^2; +b^2;=c^2;

Explanation: Ancient Chinese scholars refer to the shorter right-angled sides of right-angled triangles as "hooks", the longer right-angled sides as "strands", and the oblique sides as "strings", so this theorem becomes "pythagorean theorem". The Pythagorean Theorem reveals the relationship between the sides of a right triangle.

For example, if two right-angled sides of a right-angled triangle are 3 and 4 respectively, then the oblique side c^2= a^2+b^2=9+16=25 is c=5

Then the hypotenuse is 5.

Proof method of Pythagorean theorem

[proof method 1]

Make four congruent right-angled triangles, set their two right-angled sides to a, b, and the bevel length to c. Put them into a polygon like the one, so that D, E, F are in a straight line. The extension line of C as AC is DF at point P.

∵ D, E, F are in a straight line, and RtΔGEF ≌ RtΔEBD,

∴ ∠EGF = ∠BED,

∵ ∠EGF + ∠GEF = 90°,

∴ ∠BED + ∠GEF = 90°,

∴ ∠BEG =180o―90o= 90o.

Also ∵ AB = BE = EG = GA = c,

∴ ABEG is a square with a side length c.

∴ ∠ABC + ∠CBE = 90o.

∵ RtΔABC ≌ RtΔEBD,

∴ ∠ ABC = ∠EBD.

∴ ∠EBD + ∠CBE = 90o.

Instant ∠ CBD = 90o.

Also ∠ ∠BDE = 90o, ∠BCP = 90o,

BC = BD = a.

BD BDPC is a square with a side length a.

Similarly, HPFG is a square with side b.

Let the area of ​​the polygon GHCBE be S, then

,

Oh.

Part 3: Proof of Pythagorean Theorem

Make two congruent right-angled triangles, set their two right-angled sides to a, b, and the bevel length to c. Then make a square with side length c. Put them into a polygon as shown. Make E, A, C three points in a straight line.

Pass Q for QP∥BC, AC for point P.

Pass point B for BM⊥PQ, foot for M;

F is FN⊥PQ, and the foot is N.

∵ ∠BCA = 90o, QP∥BC,

∴ ∠MPC = 90o,

∵ BM⊥PQ,

∴ ∠BMP = 90o,

∴ BCPM is a rectangle, ie ∠MBC = 90o.

∵ ∠QBM + ∠MBA = ∠QBA = 90o,

∠ABC + ∠MBA = ∠MBC = 90o,

∴ ∠QBM = ∠ABC,

Also ∠ ∠BMP = 90o, ∠BCA = 90o, BQ = BA = c,

∴ RtΔBMQ ≌ RtΔBCA.

Similarly, RtΔQNF ≌ RtΔAEF.