Question

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Answer 1

∡MAD = 30

AD-h⇒∡ADM=90°

De aici rezulta m(∡DMA) = 90°-30° =60°

Astfel,putem aplica teorema de 30°-60°-90°, deci:

Conform T 30°-60°-90°: DM = AM/2 => AM = 10 x 2 = 20

ΔDMA-dr $T.P.\\==>$  

$AD^{2} =20^{2} -10^{2} \\AD^{2} =400-100\\AD=\sqrt{300} \\AD=10\sqrt{3}$

Fie  MF⊥AC⇒ΔMFA si ΔMFC-dr

Avem m(∡DMA)=60°⇒m(∡AMC)=180°-60°=120°

∡AMF=∡CMF=60°   (jumatate din masura ∡AMC)

ΔMFA≡ΔMFC (∡MAF=∡MCF=30°)    ⇒⇒⇒ΔMAC-isoscel

                                                                      AM=20=MC

m(∡BAD)=30°

m(∡ADC)=90°⇒m(∡ABD)=60°

AM-mediana

AM imparte BC in doua segmente egale (BM,MC)

BM=MC=20

BC=BM+MC

BC=20+20

BC=40