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Question Number 147561 by vvvv last updated on 21/Jul/21

Commented by vvvv last updated on 21/Jul/21

prove

$$\boldsymbol{{prove}} \\ $$

Answered by mr W last updated on 21/Jul/21

(m/x)=(y/n) ⇒xy=mn  (q/x)=(y/p) ⇒xy=pq    x^2 +u^2 =q^2 +w^2   n^2 +v^2 =y^2 +u^2   p^2 +w^2 =m^2 +v^2   ⇒x^2 +n^2 +p^2 +u^2 +v^2 +w^2 =y^2 +m^2 +q^2 +u^2 +v^2 +w^2   ⇒x^2 +n^2 +p^2 =y^2 +m^2 +q^2

$$\frac{{m}}{{x}}=\frac{{y}}{{n}}\:\Rightarrow{xy}={mn} \\ $$$$\frac{{q}}{{x}}=\frac{{y}}{{p}}\:\Rightarrow{xy}={pq} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{u}^{\mathrm{2}} ={q}^{\mathrm{2}} +{w}^{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} +{v}^{\mathrm{2}} ={y}^{\mathrm{2}} +{u}^{\mathrm{2}} \\ $$$${p}^{\mathrm{2}} +{w}^{\mathrm{2}} ={m}^{\mathrm{2}} +{v}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{n}^{\mathrm{2}} +{p}^{\mathrm{2}} +{u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{w}^{\mathrm{2}} ={y}^{\mathrm{2}} +{m}^{\mathrm{2}} +{q}^{\mathrm{2}} +{u}^{\mathrm{2}} +{v}^{\mathrm{2}} +{w}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} +{n}^{\mathrm{2}} +{p}^{\mathrm{2}} ={y}^{\mathrm{2}} +{m}^{\mathrm{2}} +{q}^{\mathrm{2}} \\ $$

Commented by vvvv last updated on 22/Jul/21

u=?  w=?  v=?  .....?

$$\boldsymbol{{u}}=?\:\:\boldsymbol{{w}}=?\:\:\boldsymbol{{v}}=?\:\:.....? \\ $$

Commented by mr W last updated on 22/Jul/21

u,v,w are distances from O to AC,   CB and BA.

$${u},{v},{w}\:{are}\:{distances}\:{from}\:{O}\:{to}\:{AC},\: \\ $$$${CB}\:{and}\:{BA}. \\ $$

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