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Question Number 166058 by cortano1 last updated on 12/Feb/22

Answered by som(math1967) last updated on 12/Feb/22

$$△ABC\sim △ALB\sim △CLB$$$$let\measuredangle BAC=\measuredangle DCL=\alpha$$$$AL\times AC=a^2$$$$AL=\frac{a^2}{\sqrt{a^2+b^2}}$$$$LC\times AC=b^2$$$$LC=\frac{b^2}{\sqrt{a^2+b^2}}$$$$from△ALBcos\alpha =\frac{a^2}{\sqrt{a^2+b^2}\times a}=\frac{a}{\sqrt{a^2+b^2}}$$$$from△LCD$$$$cos\alpha =\frac{{LC}^2+a^2-x^2}{2LCa}$$$$\frac{a}{\sqrt{a^2+b^2}}=\frac{\frac{b^4}{a^2+b^2}+a^2-x^2}{\frac{2b^{2}a}{\sqrt{a^2+b^2}}}$$$$\frac{2a^2{b}^2}{a^2+b^2}=\frac{b^4+a^4+a^2{b}^2}{a^2+b^2}-x^2$$$$x^2=\frac{a^4-a^2{b}^2+b^4}{a^2+b^2}=\frac{(a^2+b^2)(a^4-a^2{b}^2+b^4)}{{(a^2+b^2)}^2}$$$$x^2=\frac{(a^6+b^6)}{{(a^2+b^2)}^2}$$$$\therefore \mathit{x}=\frac{\sqrt{{\mathit{a}}^6+{\mathit{b}}^6}}{{\mathit{a}}^2+{\mathit{b}}^2}$$

Commented by som(math1967) last updated on 12/Feb/22

Commented by Tawa11 last updated on 12/Feb/22

$$\mathrm{Great}\mathrm{sir}$$

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