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Question Number 218262 by hardmath last updated on 03/Apr/25

Commented by hardmath last updated on 03/Apr/25

$$\mathrm{ABCD}-\mathrm{Rectangular}$$$$\mathrm{AK}\mathrm{\perp }\mathrm{BM}$$$${\mathrm{S}}_{△\mathbf{AKB}}={\mathrm{S}}_{△\mathbf{BMC}}={\mathrm{S}}_{\mathbf{AKMD}}$$$$\mathrm{DM}=\mathbf{a}$$$$\mathrm{Prove}\mathrm{that}:\mathrm{BC}=\mathbf{a}\sqrt{5}$$

Answered by mr W last updated on 04/Apr/25

$$sayMC=x,BC=y$$$$sayS=areaofABCD$$$$S=(a+x)y$$$$\frac{xy}{2}=\frac{S}{3}=\frac{(a+x)y}{3}$$$$\Rightarrow x=2a$$$$S_{AKB}=S_{MCB}$$$$\Rightarrow AB=MB$$$$\Rightarrow a+x=\sqrt{x^2+y^2}$$$$\Rightarrow a+2a=\sqrt{4a^2+y^2}$$$$\Rightarrow y=\sqrt{5}a✓$$

Commented by mehdee7396 last updated on 05/Apr/25

$$$$$$thankyou.iunderstand$$$$▸\mathrm{\angle }ABK=\mathrm{\angle }BMC◂$$$$$$

Commented by hardmath last updated on 04/Apr/25

$$\mathrm{thankyou}\mathrm{dear}\mathrm{professor}$$

Commented by mehdee7396 last updated on 05/Apr/25

$$$$$$\Rightarrow ?AB=MB$$

Commented by hardmath last updated on 05/Apr/25

$$\mathrm{thank}\mathrm{you}\mathrm{dear}\mathrm{professor}$$

Commented by mr W last updated on 05/Apr/25

Commented by mr W last updated on 05/Apr/25

$$ifbothtrianglesaresimilar,then$$$$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$$$$\frac{S_1}{S_2}={\left(\frac{a_1}{a_2}\right)}^2={\left(\frac{b_1}{b_2}\right)}^2={\left(\frac{c_1}{c_2}\right)}^2$$

Commented by mr W last updated on 05/Apr/25

$$\mathrm{\angle }ABK=\mathrm{\angle }MBC$$$$\Rightarrow \mathrm{\Delta }ABK\sim \mathrm{\Delta }MBC$$$$S_{\mathrm{\Delta }ABK}=S_{\mathrm{\Delta }MBC}$$$$\Rightarrow AK=BC,KB=CM,AB=BM$$$$note:iftwosimilartriangleshave$$$$equalarea,thentheyalsohaveequal$$$$sidelengthes.$$

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