电子课本网 第15页

第15页

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$65^{\circ}$
$AB// DC$
​$ (1)$​证明:∵​$AD = BE$​
∴​$AD + BD = BE + BD,$​即​$AB = DE$​
​$ $​在​$\triangle ABC$​和​$\triangle DEF $​中
​$ \begin {cases}AB = DE \\AC = DF \\BC = EF\end {cases}$​
∴​$\triangle ABC≌\triangle DEF(\mathrm {SSS})$​
​$ (2)$​解:∵​$\triangle ABC≌\triangle DEF,$​​$∠A = 55°$​
∴​$∠A = ∠F DE = 55°$​
∵​$\triangle DEF $​的内角和为​$180°,$​​$∠E = 45°$​
∴​$∠F = 180° - (∠F DE + ∠E)=180° - (55° + 45°) = 80°$​
​$ (1)$​证明:连接​$AB$​
​$ $​在​$\triangle ABC$​和​$\triangle ABD$​中
​$ \begin {cases}AC = AD \\AB = AB \\BC = BD\end {cases}$​
∴​$\triangle ABC≌\triangle ABD(\mathrm {SSS})$​
∴​$∠C = ∠D$​
​$ (2)$​解:∵​$\triangle ABC≌\triangle ABD$​
∴​$∠CAB = ∠DAB = \frac 12∠CAD,$​​$∠ABC = ∠ABD$​
∵​$∠CBD = 120°$​
∴​$∠ABC = \frac 12(360° - ∠CBD)=120°$​
∵在​$\triangle ABC$​中,​$∠C = 28°$​
∴​$∠CAB = 32°$​
∴​$∠CAD = 2∠CAB = 64°$​

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