Question

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Answer 1

Explicație pas cu pas:

notăm EG ∩ FH = {M}

$\measuredangle {EMH} = \frac{arc(EAH) + arc(FCG)}{2}</p><p> \\ = \frac{arc(EA) + arc(AH) + arc(FC) + arc(CG)}{2} \\ = \frac{ \frac{arc(AB)}{2} + \frac{arc(DA)}{2} + \frac{arc(BC)}{2} + \frac{arc(CD)}{2} }{2} \\ = \frac{arc(AB) + arc(BC) + arc(CD) + arc(DA)}{4} \\ = \frac{360 \degree}{4} = 90 \degree$

$\implies \bf EG \perp FH$