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Angelicabreaban27
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VA ROG, CAT DE REPEDE SE POATE​

VA ROG CAT DE REPEDE SE POATE class=

Răspuns :

Andyilye

Răspuns:

MD + DN = AD

Explicație pas cu pas:

[tex]Aria_{\triangle ABC} = Aria_{\triangle ADB} + Aria_{\triangle ADC} \\ \iff \frac{AD \cdot BC}{2} = \frac{MD \cdot AB}{2} + \frac{DN \cdot AC}{2} [/tex]

ΔABC este echilateral => AB ≡ AC ≡ BC

[tex]\frac{AD \cdot AB}{2} = \frac{MD \cdot AB}{2} +\frac{DN \cdot AB}{2} \\ \implies \bf AD = MD + DN[/tex]

q.e.d.

Vezi imaginea Andyilye
Targoviste44

[tex]\it AD- \hat\imath n\breve al\c{\it t}ime\ pentru\ \Delta\ echilateral \Rightarrow [AD - bisectoare \Rightarrow\\ \\ \Rightarrow \widehat{MAD}=\widehat{NAD}=\dfrac{60^o}{2}=30^o\\ \\ \\ Th.\ \angle\ 30^o\ \hat\imath n\ \Delta MAD \Rightarrow DM=\dfrac{AD}{2}\ \ \ \ \ (1)\\ \\ \\ Th.\ \angle\ 30^o\ \hat\imath n\ \Delta NAD \Rightarrow DN=\dfrac{AD}{2}\ \ \ \ \ (2)\\ \\ \\ (1),\ (2) \Rightarrow BM+DN=\dfrac{AD}{2}+\dfrac{AD}{2}=\dfrac{2AD}{2}=AD[/tex]

Vezi imaginea Targoviste44

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