10.
如图,在△ABC中,AB=AC,点D在边BC上,根据等腰三角形″三线合一″的性质填写结论:
①若BD=CD,则AD⊥BC.
②AD⊥BC,垂足为D,则BD=CD.
③若AD平分∠BAC,则AD⊥BC.
①若BD=CD,则AD⊥BC.
②AD⊥BC,垂足为D,则BD=CD.
③若AD平分∠BAC,则AD⊥BC.
7.通讯员要在规定时间内到达某地,他每小时走15千米,则可提前24分钟到达某地;如果每小时走12千米,则要迟到15分钟.设通讯员到达某地的路程是x千米,原定的时间为y小时,则可列方程组为( )
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| A. | $\left\{\begin{array}{l}{\frac{x}{15}-15=y}\\{\frac{x}{12}+12=y}\end{array}\right.$ | B. | $\left\{\begin{array}{l}{\frac{x}{15}+15=y}\\{\frac{x}{12}-12=y}\end{array}\right.$ | ||
| C. | $\left\{\begin{array}{l}{\frac{x}{15}-\frac{24}{60}=y}\\{\frac{x}{12}-\frac{15}{60}=y}\end{array}\right.$ | D. | $\left\{\begin{array}{l}{\frac{x}{15}+\frac{24}{60}=y}\\{\frac{x}{12}-\frac{15}{60}=y}\end{array}\right.$ |