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Question Number 216033 by efronzo1 last updated on 26/Jan/25

	Answered by mr W last updated on 26/Jan/25

	Commented by mr W last updated on 26/Jan/25

	$A E = B E = \frac{B C}{2}$ $A D \times {B E}^{2} + D B \times {A E}^{2} = A B \times ({D E}^{2} + A D \times B D)$ $(A D + D B) \times \frac{{B C}^{2}}{4} = A B \times ({D E}^{2} + A D \times B D)$ $\frac{{B C}^{2}}{4} = {D E}^{2} + A D \times B D$ $\Rightarrow \frac{{B C}^{2}}{A D \times B D + {D E}^{2}} = 4 ✓$
	Answered by som(math1967) last updated on 26/Jan/25

	Commented by som(math1967) last updated on 26/Jan/25

	$E L ⊥ A B$ $∴ E L = \frac{1}{2} A C, B L = A L = \frac{1}{2} A B$ ${D E}^{2} = {D L}^{2} + {E L}^{2}$ $\Rightarrow {D E}^{2} = {(B D - B L)}^{2} + \frac{{A C}^{2}}{4}$ $\Rightarrow {D E}^{2} = {B D}^{2} + {B L}^{2} - 2 B D . B L + \frac{{A C}^{2}}{4}$ $\Rightarrow {D E}^{2} = {B D}^{2} - B D . A B + \frac{{A C}^{2} + {A B}^{2}}{4}$ $\Rightarrow {D E}^{2} + B D (A B - B D) = \frac{{B C}^{2}}{4}$ $\Rightarrow {D E}^{2} + B D . A D = \frac{{B C}^{2}}{4}$ $∴ \frac{{B C}^{2}}{{D E}^{2} + B D A D} = 4$
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