Školjka

%V0 Neka je $Q \in DG \cap EF$. Neka su $\alpha, \beta = |\angle CAB|, |\angle ABC|$. Neka je $R = \frac{1}{2} |DE|$ Zbog $DE || AB$ te $DQ || AC, EQ || BC$ su $|\angle QDE|, |\angle QED| = \alpha, \beta$ pa je $\triangle ABC \sim \triangle DEQ$. Shodno tome, vrijedi $k_1 = \frac{2R}{|AB|} = \frac{|DQ|}{|AC|} = \frac{|QE|}{|BC|}$ pa vrijedi i: $$P(DEQ) = k_1^2P(ABC)$$ $$\frac{k_1|BC|\cdot|FD|}{2} = k_1^2\frac{|AB|\cdot|BC|\cdot|AC|}{4R}$$ $$|FD| = k_1\frac{|AB|\cdot|CA|}{2R}=|AC|$$ Analogno tome, vrijedi i $|GE| = |BC|$. Zbog tetivnosti $DEFG$, $|\angle GFE| = \pi - \alpha \implies |\angle QFG| = \alpha$, analogno i $|\angle QGF| = \beta$ pa vrijedi $\triangle QGF \sim \triangle ABC$ uz koeficijent $k_2=\frac{|GF|}{|AB|}$ Nadalje: $$4R^2= |EF|^2 + |FD|^2 = |DG|^2 + |GE|^2 $$ $$|BC|^2 - |AC|^2 = |EF|^2 - |DG|^2$$ $$|BC|^2 - |AC|^2 = (k_1|BC| - k_2|AC|)^2 - (k_1|AC| - k_2|BC|)^2$$ $$|BC|^2 - |AC|^2 = (k_1|BC| - k_2|AC| + k_1|AC| - k_2|BC|)(k_1|BC| - k_2|AC| - k_1|AC| + k_2|BC|)$$ $$|BC|^2 - |AC|^2 = (k_1^2 - k_2^2)(|BC|^2 - |AC|^2)$$ $$k_1^2-k_2^2 = 1$$ Ako je $\triangle ABC$ jednakokračan, vrijedi: $$|FQ|^2 + |FD|^2 = |DQ|^2$$ $$k_2^2|AC|^2 + |AC|^2 = k_1^2|AC|^2$$ $$|AC|^2 = (k_1^2 - k_2^2)|AC|^2$$ $$k_1^2-k_2^2 = 1$$ Prema tome: $P(DEFG) = P(DEQ) - P(QGF) = (k_1^2 - k_2^2)P(ABC) = P(ABC)$ $Q.E.D.$