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For your conjecture of the relationship between a square + b square and C square, choose one of them and prove it by Pythagorean theorem
The square of a plus the square of B is equal to the square of C
How to prove that the sum of squares of two short sides of an obtuse triangle is less than the square of the third side Analogy Pythagorean theorem
In the triangle ABC, the angle c is an obtuse angle
Ad is perpendicular to BC
RT triangle ADC
AD2+CD2=AC2
RT triangle in ADB
AD2+BD2=AB2
So ac2 + BC2 = (ad2 + CD2) + BC2
AB2=AD2+(BC+CD)2=AD2+CD2+BC2+2BC*CD
Because 2BC * CD is greater than 0
So AB2 is greater than ac2 + BC2
The proof method of Pythagorean theorem in right triangle
As shown in the figure, the right triangle ABC is composed of four triangles the same as triangle ABC, as shown in the figure
AB = B, AC = C, BC = a, BD = de = EF = BF = (B-A)
The area of the whole square is equal to s = C ^ 2 - (1)
The area of the whole square is equal to the area of the four triangles plus the area of the inner calibration square bdef
S=4*0.5*a*b+（b-a）^2 ——（2）
According to (1) and (2), C ^ 2 = 4 * 0.5 * a * B + (B-A) ^ 2 = a ^ 2 + B ^ 2
So C ^ 2 = a ^ 2 + B ^ 2 Pythagorean theorem is proved
In the triangle, AB is equal to 13, BC is equal to 10, and the midline ad on BC side is equal to 12. It is proved that AB is equal to AC
prove:
∵ ad is the center line of BC edge
∴BD=5
∵AB=13,AD=12,5²+12²=13²
∴∠ADB=90°
ν ad vertical bisection BC
∴AB=AC
In right triangle, according to Pythagorean theorem, the relation between acute triangle and obtuse triangle A2 + B2 and C2 is discussed Please write the process.
Acute triangle: a square + b square < C square
Obtuse triangle: a square + b square > C square
The three sides of a right triangle satisfy the Pythagorean theorem: a + B = C. what is the relationship between acute triangle and obtuse triangle? In other words, what is the relationship between a, B and C in acute and obtuse triangles?
A = B + c-2bccosa, B = a + c-2accosb, C = a + b-2abcosc
The relationship between the sum of the squares on both sides of an acute triangle and the square of the third side
The three sides are a, B, C
Then a 2 + B 2 > C
The corresponding edge of the acute triangle ABC is a B C. It is proved that a square + b square > C square Speed, hurry, please Can you tell me more about it? Anyway, thank you for answering so quickly. I don't understand
COSC = (a * a + b * B-C * c) / 2Ab > 0 leads to a * a + b * B-C * C > 0
How to prove both sides of an acute triangle if the sum of squares is greater than the third side
It can be proved directly by cosine theorem
a²=b²+c²-2bccosA
cosA=(b²+c²-a²)/2bc>0
So 0
It is proved that if a triangle is an obtuse angle triangle, the square of one side of the triangle must be greater than that of the other two sides
Any obtuse triangle
Let the long side be BC = a, the obtuse angle AC = B and Ba = C, and let the obtuse angle be a
CD is perpendicular to ab through C, ad = b * (- COSA), CD = b * (- Sina)
There are: A ^ 2 = [b * (- COSA) + C] ^ 2 + [b * (- Sina)] ^ 2
a^2=b^2+c^2^-2bccosA
Similarly:
A is an acute angle, and cosa > 0
A is a right angle, cosa = 0
A is obtuse angle, cosa < 0
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