В равнобедренном треугольнике ABC (AB=AC), в котор...

Créé le : 15 janvier 2026

Répondu en utilisant GPT-5.2 Thinking par Chat01

Question

В равнобедренном треугольнике ABC (AB=AC), в котором ∠ABC=77, отмечена середина D стороны BC. На отрезке AD выбрана точка P, а на стороне AB – точка Q так, что PQ=PC. Найдите градусную меру угла PQC.

Python

python
import sympy as sp
import math, numpy as np
h=float(math.tan(math.radians(77)))
h


4.331475874284157

Python

python
import numpy as np, math
h=math.tan(math.radians(77))
def angle_pqc(p):
    # solve intersection
    # line AB y = h*(x+1)
    # circle eq x^2 + (y-p)^2 = 1 + p^2
    # substitute:
    # x^2 + (h*(x+1)-p)^2 - (1+p^2)=0
    a=1+h**2
    b=2*h*(h-p)
    c=(h-p)**2 - (1+p**2)
    # quadratic a*x^2 + b*x + c =0 ? let's derive:
    # x^2 + (h*(x+1)-p)^2 = x^2 + (h*x + h - p)^2
    # = x^2 + (h**2*x**2 + 2*h*x*(h-p) + (h-p)**2)
    # = (1+h**2)x^2 + 2*h*(h-p)*x + (h-p)**2
    # set = 1+p^2
    # => (1+h**2)x^2 + 2*h*(h-p)*x + (h-p)**2 -1 -p^2=0
    a=1+h**2
    b=2*h*(h-p)
    c=(h-p)**2 -1 -p**2
    disc=b*b-4*a*c
    if disc< -1e-9:

Python

python
p=0.5*h
ang,Q,roots=angle_pqc(p)
ang,Q,roots


(65.21551153210612, (-1.0, 0.0), [0.05060297685041647, -1.0])

Python

python
import numpy as np, math
h=math.tan(math.radians(77))
def get_Q(p):
    a=1+h**2
    b=2*h*(h-p)
    c=(h-p)**2-1-p**2
    disc=b*b-4*a*c
    roots=[(-b+math.sqrt(disc))/(2*a),(-b-math.sqrt(disc))/(2*a)]
    # choose root not -1
    for x in roots:
        if abs(x+1)>1e-6 and -1<=x<=0:
            y=h*(x+1)
            return np.array([x,y])
    # else B
    return np.array([-1,0])
def is_cyclic(A,C,Q,P,eps=1e-6):
    # check if power of P wrt circumcircle ACQ etc? Use determinant for concyclicity
    pts=[A,C,Q,P]
    M=[]
    for x,y in pts:
        M.append([x,y,x*x+y*y,1])
    M=np.array(M,float)
    det=np.linalg.det(M)
    return det
A=np.array([0,h])
C=np.array([

Python

python
import numpy as np, math
h=math.tan(math.radians(77))
A=np.array([0,h],float);B=np.array([-1,0],float);C=np.array([1,0],float)
def ang(u,v):
    cos=np.dot(u,v)/(math.sqrt(np.dot(u,u)*np.dot(v,v)))
    cos=max(-1,min(1,cos))
    return math.degrees(math.acos(cos))
p=0.2*h
P=np.array([0,p],float); Q=get_Q(p)
# angle AQP (between QA and QP)
print("AQP", ang(A-Q,P-Q))
# angle ACQ?? Actually angle between CA and CQ at C
print("ACQ at C", ang(A-C,Q-C))
# angle APC
print("APC", ang(A-P,C-P))
# angle AQ C at Q
print("AQC", ang(A-Q,C-Q))

Python

python
def angle_at(point, p1, p2):
    # angle p1-point-p2
    import math, numpy as np
    v1=p1-point; v2=p2-point
    cos=np.dot(v1,v2)/(math.sqrt(np.dot(v1,v1)*np.dot(v2,v2)))
    cos=max(-1,min(1,cos))
    return math.degrees(math.acos(cos))
p=0.2*h
P=np.array([0,p],float); Q=get_Q(p)
print("PCQ", angle_at(C,P,Q))
print("PAQ", angle_at(A,P,Q)) # no, at A? Wait PAQ is angle between AP and AQ at A.
print("PAQ", angle_at(A,P,Q))
print("PAC", angle_at(A,P,C))

Réflexion

Confirming cyclic quadrilateral through angle equality

I’ve calculated that ∠PAQ and ∠PCQ both equal 13. Since these angles subtend the same chord PQ, it looks like quadrilateral A,P,C,Q is cyclic. To finalize this, I need to show that triangle PCQ is isosceles, so ∠PCQ also equals 13. I'll confirm this using the circle's properties.

Proving cyclicity using power and angles

I realize triangle PCQ is isosceles since PC = PQ, so base angles at C and Q are the same. This means ∠PCQ = ∠PQC, both equal to 13. To show quadrilateral A,C,Q,P is cyclic, it might be easier to prove using equal angles or power of point, not by direct angle calculation.

Exploring the cyclicity of A, C, Q, P

I’ve realized that B, C, and Q are on circle ω centered at P, which is great. But A isn’t on it. To show cyclicity of A, C, Q, P, we need A on the circle through P, C, Q — maybe using directed angles or showing equality of certain angles like ∠PAC and ∠PQC. I'll keep thinking further.

Trying to establish cyclicity of A, C, Q, P

I'm considering using cyclic quadrilateral A, C, Q, P, which would give ∠ACQ = ∠APQ. Trying to use angle chase could help, but finding a fixed approach is tricky. P being on the perpendicular bisector might be useful. Still, showing A, Q, C, P cyclic using the power of point A seems like the most straightforward path forward.

Clarifying the cyclic nature of A, C, Q, P

I’m realizing the confusion: P is the center of circle ω, but for A, C, Q, P to be cyclic, P needs to be on that circle, not its center. So, I should focus on proving that A, C, Q, P are concyclic by showing the equality of angles, like ∠PAQ = ∠PCQ, with the help of circle ω.

Relating angles in circumcircle BCQ and triangle geometry

I'm trying to connect the angle between radius CP and chord CQ with known results. The angle between the radius and chord should relate to the central angle, though it's not fixed. Using the given angle of 77° at B, I think I can relate the angles in triangle BCQ and derive the central angle at P.

Final proof for angle subtended by chord CQ

In the circumcircle with center P, angle ∠CBQ subtends arc CQ, which gives us a central angle ∠CPQ = 2 * 77 = 154°. Since triangle PCQ is isosceles, with equal base angles at C and Q, the base angles are each 13°. This confirms that the answer is 13 degrees.

Final step: ensuring Q lies on the circle and the geometry checks out

To ensure point Q lies on the circumcircle centered at P, we use the fact that PQ = PC and PB = PC, which follows from symmetry. We need to confirm P lies on the perpendicular bisector of BC, implying P is equidistant from B and C. Also, for the angle question, we assume Q ≠ B to avoid degenerate cases, ensuring a unique solution. The final angle at Q checks out as 13°.

Thought for 5m 34s

Réponse

Так как $AB=AC$ , то медиана $AD$ к основанию $BC$ является также серединным перпендикуляром к $BC$ . Значит, для любой точки $P\in AD$

PB=PC.

По условию $PQ=PC$ . Тогда вместе с $PB=PC$ получаем

PB=PC=PQ,

то есть точки $B,C,Q$ лежат на окружности с центром $P$ . Следовательно, $P$ — центр описанной окружности треугольника $BCQ$ .

Поскольку $Q\in AB$ , угол при вершине $B$ в треугольнике $BCQ$ равен

\angle CBQ=\angle CBA=\angle ABC=77^\circ.

На описанной окружности треугольника $BCQ$ центральный угол, опирающийся на хорду $CQ$ , вдвое больше вписанного угла, опирающегося на ту же хорду:

\angle CPQ = 2\angle CBQ = 2\cdot 77^\circ = 154^\circ.

В треугольнике $PCQ$ стороны $PC=PQ$ , значит он равнобедренный, и его углы при основании равны:

\angle PQC=\angle PCQ=\frac{180^\circ-154^\circ}{2}=13^\circ.

Ответ: $\boxed{13^\circ}$ .

Question

Python

Python

Python

Python

Python

Python

Réflexion

Confirming cyclic quadrilateral through angle equality

Proving cyclicity using power and angles

Exploring the cyclicity of A, C, Q, P

Trying to establish cyclicity of A, C, Q, P

Clarifying the cyclic nature of A, C, Q, P

Relating angles in circumcircle BCQ and triangle geometry

Final proof for angle subtended by chord CQ

Final step: ensuring Q lies on the circle and the geometry checks out

Réponse

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