Question-166928

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Question Number 166928 by amin96 last updated on 02/Mar/22

Answered by mr W last updated on 02/Mar/22

$${a}\:{picture}\:{says}\:{more}\:{than}\:{a}\:{thousant} \\ $$$${words}… \\ $$

Commented by mr W last updated on 03/Mar/22

Commented by mr W last updated on 03/Mar/22

ECDH is cyclic, ⇒blue angle ∠HDE=∠HCE=α similarly ⇒red angle ∠HDF=∠HBF=β ΔBFH∼ΔCEH ⇒∠HBF=∠HCE ⇒∠HDF=∠HDE

$${ECDH}\:{is}\:{cyclic},\: \\ $$$$\Rightarrow{blue}\:{angle}\:\angle{HDE}=\angle{HCE}=\alpha \\ $$$${similarly} \\ $$$$\Rightarrow{red}\:{angle}\:\angle{HDF}=\angle{HBF}=\beta \\ $$$$\Delta{BFH}\sim\Delta{CEH} \\ $$$$\Rightarrow\angle{HBF}=\angle{HCE} \\ $$$$\Rightarrow\angle{HDF}=\angle{HDE} \\ $$

Commented by som(math1967) last updated on 03/Mar/22

∠HDE=∠HCE=α ∠HBF=∠HBF=β again B,C,E,F also cyclic ∴ ∠EBF=∠ECF ∴β=α

$$\angle{HDE}=\angle{HCE}=\alpha \\ $$$$\angle{HBF}=\angle{HBF}=\beta \\ $$$${again}\:{B},{C},{E},{F}\:{also}\:{cyclic} \\ $$$$\therefore\:\angle{EBF}=\angle{ECF} \\ $$$$\therefore\beta=\alpha \\ $$$$ \\ $$

Commented by Tawa11 last updated on 03/Mar/22