(本题3分)(2019$cdot$浙江嘉兴$cdot$中考模拟)如图,点${C}$为半圆的中点,${AB}$是直径,点${D}$是半圆上一点,${AC},{BD}$交于点${E}$.若${AD}=1,{BD}=7$,则${CE}$的长为<ImageHere>
【分析】直径所对应的的圆周角为$90^{circ}$,再利用勾股定理求出${AB}$的值,然后利用${C}$点为半圆的中点判断出$triangle{ABC}$为等腰直角三角形,利用勾股定理求出${BC}$的值,最后利用三角形相似，对应边成比例求出${DE}$的长度.【详解】$because$点${C}$为半圆的中点,$therefore{AC}={BC},because{AB}$是直径,$thereforeangle{C}=angle{D}=90^{circ}$,在Rt$triangle{ADB}$中,$AD=1,BD=7,thereforeAB=5sqrt{2}$,在等腰Rt$triangleACB$中,$therefore{AC}={BC}=5,becauseangle{CBE}=angle{CAD},angle{C}=angle{D},thereforetriangle{ADE}simtriangle{BCE},thereforefrac{BC}{AD}=frac{CE}{DE}$,即$frac{5}{1}=$$frac{CE}{DE},therefore{CE}=5{DE},therefore{BE}=7-{DE}$,在Rt$triangle{CEB}$中,利用勾股定理得：$5^{2}+(5{DE})^{2}=(7-{DE})^{2}$,解得:${DE}=-frac{4}{3}$(舍去)或${DE}=frac{3}{4},quadtherefore{CE}=frac{15}{4}$故答案为$frac{15}{4}$。【点睛】直径所对的圆周角等于$90^{circ}$;两个角对应相等的三角形相似，相似三角形线段成比例;勾股定理$a^{2}+b^{2}=c^{2}$.