Question Number 204890 by mr W last updated on 29/Feb/24
Commented by mr W last updated on 29/Feb/24
$${ABCD}\:{is}\:{a}\:{square}\:{with}\:{side}\:{length}\:\mathrm{2}. \\ $$$${E}\:{is}\:{midpoint}\:{of}\:{CD},\:{F}\:{is}\:{midpoint} \\ $$$${of}\:{AD}.\:{find}\:{AP}. \\ $$
Answered by Frix last updated on 01/Mar/24
$${A}=\left(−\mathrm{1}\mid−\mathrm{1}\right),\:{B}=\left(\mathrm{1}\mid−\mathrm{1}\right),\:{C}=\left(\mathrm{1}\mid\mathrm{1}\right),\:{D}=\left(−\mathrm{1}\mid\mathrm{1}\right) \\ $$$${E}=\left(\mathrm{0}\mid\mathrm{1}\right);\:{F}=\left(−\mathrm{1}\mid\mathrm{0}\right) \\ $$$${BE}:\:{y}=−\mathrm{2}{x}+\mathrm{1};\:{CF}:\:{y}=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\:{P}=\left(\frac{\mathrm{1}}{\mathrm{5}}\mid\frac{\mathrm{3}}{\mathrm{5}}\right) \\ $$$$\Rightarrow\:\mid{AP}\mid=\mid\left(\frac{\mathrm{6}}{\mathrm{5}}\mid\frac{\mathrm{8}}{\mathrm{5}}\right)\mid=\mathrm{2} \\ $$
Commented by mr W last updated on 01/Mar/24
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Answered by mr W last updated on 01/Mar/24
Commented by mr W last updated on 01/Mar/24
$${we}\:{can}\:{easily}\:{prove}\:{that}\:{BE}\bot{CF}. \\ $$$$\Rightarrow{ABPF}\:{is}\:{cyclic}. \\ $$$$\angle{APB}=\angle{AFB}=\angle{BEC}=\angle{ABP} \\ $$$$\Rightarrow\Delta{ABP}\:{is}\:{isosceles}. \\ $$$$\Rightarrow{AP}={AB}=\mathrm{2}\:\checkmark \\ $$