In the triangle B2 + C2 BC = A2, find a (ABC followed by square)

In the triangle B2 + C2 BC = A2, find a (ABC followed by square)

Cosa = (b ^ 2 + C ^ 2-A ^ 2) / 2BC = BC / 2BC = 1 / 2, so a = 60 degrees

As shown in the figure, in △ ABC, let the lengths of BC, Ca and ab be a, B and C respectively, and prove that A2 = B2 + c2-2bccosa

It is known that the opposite sides of a, B and C in △ ABC are a, B and C respectively. The Cartesian coordinate system is established with a as the origin and ab as the x-axis, then C (bcosa, bsina), B (C, 0), (A2 = | BC | 2 = (bcosa-c) 2 + (bsina) 2 = b2cos2a-2bccosa + C2 + b2sin2a = B2 + c2-2bccosa

In the triangle ABC, angle B = 60 degrees, a = 4, angle a = 45 degrees, find a2-ac + c2-b2, (2 is the square) for detailed explanation,

Cosine theorem: 2BC cosa = B2 + c2-a2; 2Ac CoSb = A2 + c2-b2
a2-ac+c2-b2=2ac cosB-ac=2ac*1/2-ac=0

It is known that the isosceles right triangle ABC. D is BC, the midpoint ad is vertical CE, and the point E is on ab. it is proved that the angle CDA is equal to the angle EDB

Let CF ⊥ AB be f,
Then ∠ ACF = 45 °,
In △ ABC, ∠ ACB = 90 ° CE ⊥ ad,
Thus, from ∠ ACG ＝ B = 45 ° AB = AC,
It is easy to prove that ∠ 1 = 2,
Thus △ AGC ≌ △ CEB (ASA) can be obtained
Then CD = dB, CG = be, ∠ GCD = B,
In addition, △ CGD ≌ △ bed (SAS) can be obtained,
Then it can be proved that CDA = EDB