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Question Number 109591 by 1549442205PVT last updated on 25/Aug/20

An isosceles AEF is inscribed into a  square ABCD such that pointE is on  side BC,point F is on side CDand AE=EF.  Knownthat tanAEF^(�) =2.Find tanFEC^(�)

$$\mathrm{An}\:\mathrm{isosceles}\:\mathrm{AEF}\:\mathrm{is}\:\mathrm{inscribed}\:\mathrm{into}\:\mathrm{a} \\ $$$$\mathrm{square}\:\mathrm{ABCD}\:\mathrm{such}\:\mathrm{that}\:\mathrm{pointE}\:\mathrm{is}\:\mathrm{on} \\ $$$$\mathrm{side}\:\mathrm{BC},\mathrm{point}\:\mathrm{F}\:\mathrm{is}\:\mathrm{on}\:\mathrm{side}\:\mathrm{CDand}\:\mathrm{AE}=\mathrm{EF}. \\ $$$$\mathrm{Knownthat}\:\mathrm{tan}\widehat {\mathrm{AEF}}=\mathrm{2}.\mathrm{Find}\:\mathrm{tan}\widehat {\mathrm{FEC}} \\ $$

Commented by 1549442205PVT last updated on 24/Aug/20

Commented by mr W last updated on 24/Aug/20

if AE=AF, then tan ∠FEC=tan 45°=1

$${if}\:{AE}={AF},\:{then}\:\mathrm{tan}\:\angle{FEC}=\mathrm{tan}\:\mathrm{45}°=\mathrm{1} \\ $$

Commented by 1549442205PVT last updated on 25/Aug/20

That is correct.But here AE=EF

$$\mathrm{That}\:\mathrm{is}\:\mathrm{correct}.\mathrm{But}\:\mathrm{here}\:\mathrm{AE}=\mathrm{EF} \\ $$

Answered by mr W last updated on 24/Aug/20

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