Cap. 7 - Cuadriláteros ⇒ Solucionário:Racso - Cap VII - Problemas de Geometria y Como Resolverlos - I Edição - Ex:-02 Tópico resolvido

[petras]

Problema Proposto
02 - As mediatrizes dos lados AD e CD de um paralelogramo ABCD se interceptam em
um ponto M que pertence a BC. Achar a [tex3]\mathsf{\measuredangle MAD}[/tex3] se [tex3]\mathsf{\measuredangle B=110^\circ}[/tex3].

Resposta

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C) 40^o

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[tex3]\triangle MGC \sim \triangle MGD(LAL) = 90^\circ\\
\measuredangle MGC = 180^\circ - 110^\circ = 70^\circ \therefore \measuredangle MGC = \measuredangle MDG = 70^\circ\\
\measuredangle D = \measuredangle B = 110^0\\
\therefore \measuredangle MDF = \measuredangle CDA - \measuredangle MDG=110^\circ - 70^\circ \therefore \boxed{\color{red}\measuredangle MAD = 40^\circ}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\triangle MGC \sim \triangle MGD(LAL) = 90^\circ\\
\measuredangle MGC = 180^\circ - 110^\circ = 70^\circ \therefore \measuredangle MGC = \measuredangle MDG = 70^\circ\\
\measuredangle D = \measuredangle B = 110^0\\
\therefore \measuredangle MDF = \measuredangle CDA - \measuredangle MDG=110^\circ - 70^\circ \therefore \boxed{\color{red}\measuredangle MAD = 40^\circ}[/tex3]

(Solução: JohnOmielan)

[petras]

Outra resoluçao:
[tex3]110^{\circ}=\measuredangle GDA=\measuredangle MDA+\measuredangle MDG (I)\\
mas ~\measuredangle MDG=90^{\circ}-\measuredangle DMG=90^{\circ}-\frac{1}{2}\measuredangle CMD=90^{\circ}-\frac{1}{2}\measuredangle MDA=90^{\circ}-\frac{x}{2} \\
Substituindo ~em ~(I): 110 = x+90^{\circ}-\frac{x}{2}\\
\therefore \boxed{\color{red}x = 40^o} [/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]110^{\circ}=\measuredangle GDA=\measuredangle MDA+\measuredangle MDG (I)\\
mas ~\measuredangle MDG=90^{\circ}-\measuredangle DMG=90^{\circ}-\frac{1}{2}\measuredangle CMD=90^{\circ}-\frac{1}{2}\measuredangle MDA=90^{\circ}-\frac{x}{2} \\
Substituindo ~em ~(I): 110 = x+90^{\circ}-\frac{x}{2}\\
\therefore \boxed{\color{red}x = 40^o} [/tex3]

(Solução: Michael)