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In-a-rectangle-ABCD-E-is-the-midpoint-of-AB-F-is-a-point-on-AC-such-that-BF-is-perpendicular-to-AC-and-FE-perpendicular-to-BD-Suppose-BC-8-3-Find-AB-




Question Number 20599 by Tinkutara last updated on 28/Aug/17
In a rectangle ABCD, E is the midpoint  of AB; F is a point on AC such that BF  is perpendicular to AC; and FE  perpendicular to BD. Suppose BC = 8(√3).  Find AB.
$$\mathrm{In}\:\mathrm{a}\:\mathrm{rectangle}\:{ABCD},\:{E}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint} \\ $$$$\mathrm{of}\:{AB};\:{F}\:\mathrm{is}\:\mathrm{a}\:\mathrm{point}\:\mathrm{on}\:{AC}\:\mathrm{such}\:\mathrm{that}\:{BF} \\ $$$$\mathrm{is}\:\mathrm{perpendicular}\:\mathrm{to}\:{AC};\:\mathrm{and}\:{FE} \\ $$$$\mathrm{perpendicular}\:\mathrm{to}\:{BD}.\:\mathrm{Suppose}\:{BC}\:=\:\mathrm{8}\sqrt{\mathrm{3}}. \\ $$$$\mathrm{Find}\:{AB}. \\ $$
Answered by ajfour last updated on 29/Aug/17
Commented by Tinkutara last updated on 29/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 29/Aug/17
EF=AE=BE (radii of circle  with E as centre)  so ∠AFE=θ  ∠DBF=90°−2θ , so ∠EBF=2θ  now ∠AFE+∠EFB=90°  so              θ+2θ=90°    ⇒      tan θ=((BC)/(AB)) =((8(√3))/(AB)) = (1/( (√3)))      or   AB=24 .
$${EF}={AE}={BE}\:\left({radii}\:{of}\:{circle}\right. \\ $$$$\left.{with}\:{E}\:{as}\:{centre}\right) \\ $$$${so}\:\angle{AFE}=\theta \\ $$$$\angle{DBF}=\mathrm{90}°−\mathrm{2}\theta\:,\:{so}\:\angle{EBF}=\mathrm{2}\theta \\ $$$${now}\:\angle{AFE}+\angle{EFB}=\mathrm{90}° \\ $$$${so}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\theta+\mathrm{2}\theta=\mathrm{90}°\:\: \\ $$$$\Rightarrow\:\:\:\:\:\:\mathrm{tan}\:\theta=\frac{{BC}}{{AB}}\:=\frac{\mathrm{8}\sqrt{\mathrm{3}}}{{AB}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:{or}\:\:\:\boldsymbol{{AB}}=\mathrm{24}\:. \\ $$$$ \\ $$

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