Question-220630

Question Number 220630 by Mingma last updated on 17/May/25

Answered by mr W last updated on 17/May/25

Commented by mr W last updated on 17/May/25

ΔBCD is equilateral. ⇒BD=BC=CD=a, say say the center of circumcircle from ΔABD is at O. we have ∠BOD=2×∠BAD=2×30°=60° ⇒OA=OB=OD=radius=BD=a acc. to cosine law: AB^2 =a^2 +a^2 −2a×a cos θ =2a^2 −2a^2 cos θ AD^2 =a^2 +a^2 −2a×a cos (2π−θ−(π/3)) =2a^2 −2a^2 cos (θ+(π/3)) AB^2 +AD^2 =4a^2 −(√3)a^2 ((√3) cos θ−sin θ) acc. to cosine law again: AC^2 =a^2 +((√3)a)^2 −2(√3)a^2 cos (θ+(π/6)) =4a^2 −(√3)a^2 ((√3) cos θ−sin θ) ⇒AC^2 =AB^2 +AD^2

$$\Delta{BCD}\:{is}\:{equilateral}. \\ $$$$\Rightarrow{BD}={BC}={CD}={a},\:{say} \\ $$$${say}\:{the}\:{center}\:{of}\:{circumcircle}\:{from} \\ $$$$\Delta{ABD}\:{is}\:{at}\:{O}.\:{we}\:{have}\: \\ $$$$\angle{BOD}=\mathrm{2}×\angle{BAD}=\mathrm{2}×\mathrm{30}°=\mathrm{60}° \\ $$$$\Rightarrow{OA}={OB}={OD}={radius}={BD}={a} \\ $$$${acc}.\:{to}\:{cosine}\:{law}: \\ $$$${AB}^{\mathrm{2}} ={a}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{a}×{a}\:\mathrm{cos}\:\theta \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} \:\mathrm{cos}\:\theta \\ $$$${AD}^{\mathrm{2}} ={a}^{\mathrm{2}} +{a}^{\mathrm{2}} −\mathrm{2}{a}×{a}\:\mathrm{cos}\:\left(\mathrm{2}\pi−\theta−\frac{\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} \:\mathrm{cos}\:\left(\theta+\frac{\pi}{\mathrm{3}}\right) \\ $$$${AB}^{\mathrm{2}} +{AD}^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} −\sqrt{\mathrm{3}}{a}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right) \\ $$$${acc}.\:{to}\:{cosine}\:{law}\:{again}: \\ $$$${AC}^{\mathrm{2}} ={a}^{\mathrm{2}} +\left(\sqrt{\mathrm{3}}{a}\right)^{\mathrm{2}} −\mathrm{2}\sqrt{\mathrm{3}}{a}^{\mathrm{2}} \:\mathrm{cos}\:\left(\theta+\frac{\pi}{\mathrm{6}}\right) \\ $$$$\:\:\:=\mathrm{4}{a}^{\mathrm{2}} −\sqrt{\mathrm{3}}{a}^{\mathrm{2}} \left(\sqrt{\mathrm{3}}\:\mathrm{cos}\:\theta−\mathrm{sin}\:\theta\right) \\ $$$$\Rightarrow{AC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AD}^{\mathrm{2}} \\ $$

Commented by Mingma last updated on 17/May/25

Can you please show details of your algebra? Also if you can use vector method as well by putting it in x-y plane would be nice. I would love to see more analytical way for this problem

Commented by mr W last updated on 17/May/25

$${i}'{ve}\:{added}\:{some}\:{explanation}.\:{if}\:{you} \\ $$$${need}\:{more},\:{please}\:{tell}\:{me}\:{where}. \\ $$

Commented by mr W last updated on 18/May/25

$${a}\:{pure}\:{geometry}\:{method}\:{see}\:{Q}\mathrm{220726} \\ $$

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