Question-199688

Question Number 199688 by cortano12 last updated on 07/Nov/23

Commented by cortano12 last updated on 07/Nov/23

$$\mathrm{prove}\:\mathrm{that}\:\mathrm{formula} \\ $$

Answered by AST last updated on 07/Nov/23

Let d and e be the diagonals of the cyclic quadrilateral; Ptolemy′s theorem⇒de=2br+ac...(i) Also, d=(√(4r^2 −a^2 ));e=(√(4r^2 −c^2 )) ⇒de=(√(16r^4 −4r^2 c^2 −4a^2 r^2 +a^2 c^2 ))=2br+ac ⇒16r^4 −4r^2 c^2 −4a^2 r^2 +a^2 c^2 =4b^2 r^2 +a^2 c^2 +4abcr ⇒4r^2 (4r^2 −c^2 −a^2 )=4r^2 (b^2 +((abc)/r)) ⇒a^2 +b^2 +c^2 +((abc)/r)=4r^2

$${Let}\:{d}\:{and}\:{e}\:{be}\:{the}\:{diagonals}\:{of}\:{the}\:{cyclic}\: \\ $$$${quadrilateral}; \\ $$$${Ptolemy}'{s}\:{theorem}\Rightarrow{de}=\mathrm{2}{br}+{ac}…\left({i}\right) \\ $$$${Also},\:{d}=\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} };{e}=\sqrt{\mathrm{4}{r}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$$$\Rightarrow{de}=\sqrt{\mathrm{16}{r}^{\mathrm{4}} −\mathrm{4}{r}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} }=\mathrm{2}{br}+{ac} \\ $$$$\Rightarrow\mathrm{16}{r}^{\mathrm{4}} −\mathrm{4}{r}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{4}{a}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} =\mathrm{4}{b}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} +\mathrm{4}{abcr} \\ $$$$\Rightarrow\mathrm{4}{r}^{\mathrm{2}} \left(\mathrm{4}{r}^{\mathrm{2}} −{c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)=\mathrm{4}{r}^{\mathrm{2}} \left({b}^{\mathrm{2}} +\frac{{abc}}{{r}}\right) \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\frac{{abc}}{{r}}=\mathrm{4}{r}^{\mathrm{2}} \\ $$

Answered by mr W last updated on 07/Nov/23

Commented by mr W last updated on 07/Nov/23

β=π−α ⇒cos β=−cos α=−(c/(2r)) a^2 +b^2 +2ab×(c/(2r))=AC^2 =(2r)^2 −c^2 ⇒a^2 +b^2 +c^2 +((abc)/r)=4r^2

$$\beta=\pi−\alpha\:\Rightarrow\mathrm{cos}\:\beta=−\mathrm{cos}\:\alpha=−\frac{{c}}{\mathrm{2}{r}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ab}×\frac{{c}}{\mathrm{2}{r}}={AC}^{\mathrm{2}} =\left(\mathrm{2}{r}\right)^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\frac{{abc}}{{r}}=\mathrm{4}{r}^{\mathrm{2}} \\ $$

Answered by som(math1967) last updated on 07/Nov/23

cos∠BAC=(a/(2r)) cos(180−∠BAC)=((b^2 +c^2 −BD^2 )/(2bc)) (a/(2r))=((BD^2 −b^2 −c^2 )/(2bc)) ⇒(a/r)=((4r^2 −a^2 −b^2 −c^2 )/(bc)) ⇒((abc)/r) +a^2 +b^2 +c^2 =4r^2

$${cos}\angle{BAC}=\frac{{a}}{\mathrm{2}{r}} \\ $$$${cos}\left(\mathrm{180}−\angle{BAC}\right)=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{BD}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\:\frac{{a}}{\mathrm{2}{r}}=\frac{{BD}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{\mathrm{2}{bc}} \\ $$$$\Rightarrow\frac{{a}}{{r}}=\frac{\mathrm{4}{r}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{bc}} \\ $$$$\Rightarrow\frac{{abc}}{{r}}\:+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} \\ $$

Commented by som(math1967) last updated on 07/Nov/23

Answered by MM42 last updated on 07/Nov/23

AB^2 =a^2 +b^2 −2abcosα AB^2 =4r^2 −c^2 ⇒a^2 +b^2 +c^2 −2abcosα=4r^2 ⇒a^2 +b^2 +c^2 −2abcos(π−β)=4r^2 ⇒a^2 +b^2 +c^2 +2abcosβ=4r^2 ⇒a^2 +b^2 +c^2 +((abc)/r)=4r^2 ✓

$${AB}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{abcos}\alpha \\ $$$${AB}^{\mathrm{2}} =\mathrm{4}{r}^{\mathrm{2}} −{c}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{abcos}\alpha=\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{abcos}\left(\pi−\beta\right)=\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{abcos}\beta=\mathrm{4}{r}^{\mathrm{2}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\frac{{abc}}{{r}}=\mathrm{4}{r}^{\mathrm{2}} \:\:\checkmark \\ $$$$ \\ $$

Commented by MM42 last updated on 07/Nov/23

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