方べきの定理
$PA・PB=PC・PD$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/88c2274b58836af473d6e265b574e0bc-2.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/0c03f0db56c7367d7e9e288a9b7bd609.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/d5b01c74591b0ba3d2fc7fe2b8a840e9.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/03/51bae72de686cf9eec05d28c967f24d3.jpg)
簡単にまとめるとこんな感じ!
$□\cdot□$$ = $$□\cdot□$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/7b77708a0815bfa425735d380ab62f5c.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/7e5032353069b61783c20fa6ab96a992-1.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/54f92786ef7a60b820b71007733e0ac8-1.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/03/949e0b080dc0cd3c8380884ac56b6c57.jpg)
方べきの定理はこの3種類を使いこなせばいいんだね!
証明
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/88c2274b58836af473d6e265b574e0bc-3.png)
[証明]
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/f5084c274dbd3badc361c33312ada779-2.png)
$\triangle PAC$ と $\triangle PDB$ を考える
円周角の定理より
$\angle PAC=\angle PDB$
$\angle PCA=\angle PBD$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/d2d802f68b21150d40fb635727c7153b.png)
2組の角がそれぞれ等しいので
$\triangle PAC\backsim\triangle PDB$
したがって
$PA:PC=PD:PB$
これより
$PA・PB=PC・PD$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/0c03f0db56c7367d7e9e288a9b7bd609-1.png)
[証明]
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/041b8b9bc605736b54791ea08cf6e17e.png)
$\triangle PAC$ と $\triangle PDB$ を考える
円に内接する四角形の性質より
$\angle PAC=\angle PDB$
$\angle PCA=\angle PBD$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/10e958dbd007ce3621caff3933bd57d0.png)
2組の角がそれぞれ等しいので
$\triangle PAC\backsim\triangle PDB$
したがって
$PA:PC=PD:PB$
これより
$PA・PB=PC・PD$
円に内接する四角形の性質はこれ↓
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/20210605-160x90.jpg)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/d5b01c74591b0ba3d2fc7fe2b8a840e9-1.png)
[証明]
下図のように円の中心 $O$,点 $C$,$D$ をとる
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/49fa0437f3ff0ff6dc01e112b9d9e5de-1.png)
方べきの定理より
$PA・PB=PC・PD$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/0f355c23fd177082e7973aac60c8ee6c.png)
円の半径を $r$ とすると
$PC=PO-r$,$PD=PO+r$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/ac8cbdb5ebb9c0a17f60f742cac781ee-1.png)
これより
$PA・PB=PC・PD=(PO-r)(PO+r)$
よって
$PA・PB=PO^2-r^2$
$\triangle OPT$ において三平方の定理より
$PT^2=PO^2-r^2$
したがって
$PA・PB=PT^2$
まとめ
● 方べきの定理
$□・□$$ = $$□・□$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/7b77708a0815bfa425735d380ab62f5c.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/7e5032353069b61783c20fa6ab96a992-1.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/54f92786ef7a60b820b71007733e0ac8-1.png)
問題
(1)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/32167ef1e804b3b156056d95f34af47d.png)
方べきの定理より
$x・2=3・4$
$x=6$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/9aa1f9659e205a079f77e41f4b03c874.png)
(2)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/d51d6e5938159c90736ebf7ece31f9c5.png)
方べきの定理より
$x(x+2)=3・8$
$x^2+2x=24$
$x^2+2x-24=0$
$(x-4)(x+6)=0$
$x>0$ より $x=4$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/a83acc74ce9eb7f94ef8a95557085bde.png)
(3)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/8aeaed729f374004219bb5d182156bcb.png)
方べきの定理より
$x^2=2・6$
$x^2=12$
$x>0$ より $x=2\sqrt{3}$
![](https://enjoy-mathematics.com/wp-content/uploads/2021/06/469ec66286b0a77bcc67f78db20749d6.png)
![](https://enjoy-mathematics.com/wp-content/uploads/2021/03/51bae72de686cf9eec05d28c967f24d3.jpg)
3つの方べきの定理を使いこなそう!
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