{
  "aliases": null,
  "backlinks": null,
  "body": "\n## Proof\n\nWe compare both angles with the angle at the centre.\n\n1. The points are [[concyclic|049DCJZ448]] in $\\mathbb{R}^2$, so they lie on a [[circle|060W5CBZP6]] with a centre $O$ and a radius $r \\geq 0$. Since $A \\neq B$, we have $r \u003e 0$.\n2. By the [[inscribed angle theorem|0GA1V8NN8C]] with $V = C$ and with $V = D$, we get $2\\,\\measuredangle ACB = \\measuredangle AOB = 2\\,\\measuredangle ADB$. This is the first claim.\n3. Let $\\theta_C$ and $\\theta_D$ be real numbers that represent the [[oriented angles|03YN70NK0Z]] $\\measuredangle ACB$ and $\\measuredangle ADB$. By step 2, $2\\theta_C - 2\\theta_D \\in 2\\pi\\mathbb{Z}$, so $\\theta_C - \\theta_D \\in \\pi\\mathbb{Z}$.\n4. Hence either $\\theta_C \\equiv \\theta_D$ or $\\theta_C \\equiv \\theta_D + \\pi$ modulo $2\\pi$.\n5. For the second part, put $f(X) = \\det(B - A, X - A)$ for $X \\in \\mathbb{R}^2$, with the [[determinant|00KDZRW4C0]]. Since $B - A \\neq 0$, $f(X) = 0$ if and only if $X$ is on the [[line|0MP9BCF6DP]] $AB$.\n6. The points of the line $AB$ are $A + t(B - A)$. Such a point is on the circle when $\\norm{A - O + t(B - A)}^2 = r^2$. This is a quadratic equation in $t$ with the leading coefficient $\\norm{B - A}^2 \u003e 0$.\n7. So the line meets the circle in at most two points, namely $A$ and $B$. Hence $C$ and $D$ are not on the line $AB$, and $f(C) \\neq 0$, $f(D) \\neq 0$.\n8. Let $X$, $Y$ be points off the line $AB$. They are on the same side of it when the [[segment|VQ4Q2V45RP]] $XY$ misses the line. The determinant is linear in its second column, so $f((1 - t)X + tY) = (1 - t)f(X) + t f(Y)$.\n9. If $f(X)$ and $f(Y)$ have the same sign, then $(1 - t)f(X) + tf(Y)$ has that sign for $0 \\leq t \\leq 1$. So the segment misses the line.\n10. If they have opposite signs, then $t = \\frac{f(X)}{f(X) - f(Y)}$ lies in $(0, 1)$ and makes the value zero. So the segment meets the line. Thus $X$, $Y$ are on the same side if and only if $f(X) f(Y) \u003e 0$.\n11. By the definition of the oriented angle, $\\sin\\theta_C = \\frac{\\det(A - C, B - C)}{\\norm{A - C}\\norm{B - C}}$.\n12. Since $B - C = (B - A) + (A - C)$, we get $\\det(A - C, B - C) = \\det(A - C, B - A) = \\det(B - A, C - A) = f(C)$. So $\\sin\\theta_C$ has the sign of $f(C)$. In the same way, $\\sin\\theta_D$ has the sign of $f(D)$.\n13. The [[angle|0GJAJC6TNZ]] $\\angle ACB$ is the unique number in $[0, \\pi]$ whose cosine is $\\cos\\theta_C$, since both equal $\\frac{(A - C) \\cdot (B - C)}{\\norm{A - C}\\norm{B - C}}$. The same holds for $\\angle ADB$ and $\\theta_D$.\n14. Suppose that $C$ and $D$ are on the same side. Then $\\sin\\theta_C$ and $\\sin\\theta_D$ have the same sign by steps 10 and 12. Since [[sine has period $2\\pi$|0P4NQD746K]] and $\\sin(\\theta + \\pi) = -\\sin\\theta$, the case $\\theta_C \\equiv \\theta_D + \\pi$ is excluded.\n15. So $\\theta_C \\equiv \\theta_D$, hence $\\cos\\theta_C = \\cos\\theta_D$. By step 13, $\\angle ACB = \\angle ADB$.\n16. Suppose that $C$ and $D$ are on opposite sides. Then the sines have opposite signs, so $\\theta_C \\not\\equiv \\theta_D$. By step 4, $\\theta_C \\equiv \\theta_D + \\pi$, so $\\cos\\theta_C = -\\cos\\theta_D$.\n17. The number $\\pi - \\angle ADB$ lies in $[0, \\pi]$, and its cosine is $-\\cos\\angle ADB = -\\cos\\theta_D = \\cos\\theta_C$. By step 13, $\\angle ACB = \\pi - \\angle ADB$, that is, $\\angle ACB + \\angle ADB = \\pi$.\n\n## Idea\n\nBoth inscribed angles are half the central angle, but only up to a half turn. The sign of the sine of an oriented angle tells on which side of $AB$ the vertex lies. On the same side the half turn cannot appear, and on opposite sides it must appear.\n",
  "commit": "4a3290d299c8b2cd6493fce2296920ba7f82d942",
  "front": {
    "kind": "proof",
    "msc": [
      "51M04"
    ],
    "name": "Proof that angles in the same segment are equal",
    "proves": "0GS0AW6JRY",
    "sources": [
      {
        "locator": "Version relative aux angles orientés de droites",
        "url": "https://fr.wikipedia.org/wiki/Th%C3%A9or%C3%A8me_de_l%27angle_inscrit_et_de_l%27angle_au_centre"
      }
    ],
    "tag": "PRPWTE5WH6",
    "uses": [
      "0GA1V8NN8C",
      "049DCJZ448",
      "03YN70NK0Z",
      "0GJAJC6TNZ",
      "060W5CBZP6",
      "0MP9BCF6DP",
      "VQ4Q2V45RP",
      "00KDZRW4C0",
      "0P4NQD746K"
    ]
  },
  "kind": "proof",
  "link": "https://math2.tamnd.com/node/PRPWTE5WH6/",
  "links": [
    {
      "field": "proves",
      "target": "0GS0AW6JRY"
    },
    {
      "field": "uses",
      "target": "0GA1V8NN8C"
    },
    {
      "field": "uses",
      "target": "049DCJZ448"
    },
    {
      "field": "uses",
      "target": "03YN70NK0Z"
    },
    {
      "field": "uses",
      "target": "0GJAJC6TNZ"
    },
    {
      "field": "uses",
      "target": "060W5CBZP6"
    },
    {
      "field": "uses",
      "target": "0MP9BCF6DP"
    },
    {
      "field": "uses",
      "target": "VQ4Q2V45RP"
    },
    {
      "field": "uses",
      "target": "00KDZRW4C0"
    },
    {
      "field": "uses",
      "target": "0P4NQD746K"
    }
  ],
  "markdown": "https://math2.tamnd.com/node/PRPWTE5WH6.md",
  "msc": [
    "51M04"
  ],
  "name": "Proof that angles in the same segment are equal",
  "prov": null,
  "role": "",
  "source": "https://github.com/tamnd/math2/blob/main/nodes/PRP/PRPWTE5WH6.md",
  "status": null,
  "tag": "PRPWTE5WH6",
  "uri": "m2://PRPWTE5WH6",
  "url": "https://math2.tamnd.com/wiki/proof-that-angles-in-the-same-segment-are-equal/",
  "writing": {
    "tag": "PRPWTE5WH6",
    "kind": "proof",
    "level": "W0",
    "next": [
      {
        "what": "an inventory row",
        "note": "the row of the inventory that the node accounts for",
        "have": 0,
        "want": 1
      }
    ]
  }
}