Montrer que: 4r^2 =a^2 +b^2 +c^2 +d^2


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Question Number 187142 by a.lgnaoui last updated on 14/Feb/23

$M o n t r e r q u e :$ $4 r^{2} = a^{2} + b^{2} + c^{2} + d^{2}$
Commented by a.lgnaoui last updated on 14/Feb/23

Commented by mahdipoor last updated on 14/Feb/23
$C(O,r) : x^2 +y^2 =r^2 P=(i,j) { ((line CPD : x=i)),((line APB : y=j)) :} A=(−(√(r^2 −j^2 )),j) B=(+(√(r^2 −j^2 )),j) C=(i,(√(r^2 −i^2 ))) D=(i,−(√(r^2 −i^2 ))) a^2 +b^2 +c^2 +d^2 =∣AP∣^2 +∣AB∣^2 +...= [(i−(−(√(r^2 −j^2 ))))^2 +(j−j)^2 ]+[(i−(√(r^2 −j^2 )))^2 +(j−j)^2 ] +...=2i^2 +2(r^2 −j^2 )+2j^2 +2(r^2 −i^2 )=4r^2$
$C (O, r) : x^{2} + y^{2} = r^{2}$ $P = (i, j) {\begin{cases} l i n e C P D : x = i \\ l i n e A P B : y = j \end{cases}$ $A = (- \sqrt{r^{2} - j^{2}}, j) B = (+ \sqrt{r^{2} - j^{2}}, j)$ $C = (i, \sqrt{r^{2} - i^{2}}) D = (i, - \sqrt{r^{2} - i^{2}})$ $a^{2} + b^{2} + c^{2} + d^{2} =∣ A P ∣^{2} + ∣ A B ∣^{2} + . . . =$ $[{(i - (- \sqrt{r^{2} - j^{2}}))}^{2} + {(j - j)}^{2}] + [{(i - \sqrt{r^{2} - j^{2}})}^{2} + {(j - j)}^{2}]$ $+ . . . = 2 i^{2} + 2 (r^{2} - j^{2}) + 2 j^{2} + 2 (r^{2} - i^{2}) = 4 r^{2}$
	Answered by mr W last updated on 14/Feb/23

	${(b - \frac{a + b}{2})}^{2} + {(\frac{c + d}{2})}^{2} = r^{2}$ $\Rightarrow {(a - b)}^{2} + {(c + d)}^{2} = 4 r^{2} . . . (i)$ ${(d - \frac{c + d}{2})}^{2} + {(\frac{a + b}{2})}^{2} = r^{2}$ $\Rightarrow {(c - d)}^{2} + {(a + b)}^{2} = 4 r^{2} . . . (i i)$ $(i) - (i i) :$ $4 c d - 4 a b = 0 \Rightarrow c d = a b$ $(i) :$ $a^{2} + b^{2} - 2 a b + c^{2} + d^{2} + 2 c d = 4 r^{2}$ $\Rightarrow a^{2} + b^{2} + c^{2} + d^{2} = 4 r^{2} ✓$
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