《2025年假日数学吉林出版集团股份有限公司八年级人教版》

15. 如图,在四边形ABCD中,$AC\perp BD$,$BD = 12$,$AC = 16$,E,F分别为AB,CD的中点,求EF的长.

10

答案： 取 $BC$ 边的中点 $G$，连接 $EG$，$FG$。$ \because E$，$F$ 分别为 $AB$，$CD$ 的中点，$ \therefore EG$ 是 $ \triangle ABC$ 的中位线，$FG$ 是 $ \triangle BCD$ 的中位线。$ \therefore EG \stackrel { // } { = } \frac { 1 } { 2 }AC$，$FG \stackrel { // } { = } \frac { 1 } { 2 }BD$。又 $ \because BD = 12$，$AC = 16$，$AC \perp BD$，$ \therefore EG = 8$，$FG = 6$，$EG \perp FG$。$ \therefore $ 在 $ \mathrm { Rt } \triangle EGF$ 中，由勾股定理，得 $EF = \sqrt { EG ^ { 2 } + FG ^ { 2 } } = \sqrt { 8 ^ { 2 } + 6 ^ { 2 } } = 10$，即 $EF$ 的长度是 10。

16. 如图,在四边形ABCD中,$\angle ABC = 90^{\circ}$,E是AC的中点,且$AC = AD$.
(1)尺规作图：作$\angle CAD$的平分线AF,交CD于点F,连接EF,BF(保留作图痕迹,不写作法)；
略
(2)在(1)所作的图中,若$\angle BAD = 45^{\circ}$,且$\angle CAD = 2\angle BAC$,求证：$\triangle BEF$是等边三角形.
$ \because AC = AD$，$AF$ 平分 $ \angle CAD$，$ \therefore \angle CAF = \angle DAF$，$AF \perp CD$。$ \because \angle CAD = 2 \angle BAC$，$ \angle BAD = 45^{\circ}$，$ \therefore \angle BAE = \angle EAF = \angle FAD = 15^{\circ}$。$ \because \angle ABC = \angle AFC = 90^{\circ}$，$AE = EC$，$ \therefore BE = AE = EC$，$EF = AE = EC$。$ \therefore EB = EF$，$ \angle EAB = \angle EBA = 15^{\circ}$，$ \angle EAF = \angle EFA = 15^{\circ}$。$ \therefore \angle BEC = \angle EAB + \angle EBA = 30^{\circ}$，$ \angle CEF = \angle EAF + \angle EFA = 30^{\circ}$。$ \therefore \angle BEF = 60^{\circ}$。$ \therefore \triangle BEF$ 是等边三角形。

答案：

(1) 略
(2) $ \because AC = AD$，$AF$ 平分 $ \angle CAD$，$ \therefore \angle CAF = \angle DAF$，$AF \perp CD$。$ \because \angle CAD = 2 \angle BAC$，$ \angle BAD = 45^{\circ}$，$ \therefore \angle BAE = \angle EAF = \angle FAD = 15^{\circ}$。$ \because \angle ABC = \angle AFC = 90^{\circ}$，$AE = EC$，$ \therefore BE = AE = EC$，$EF = AE = EC$。$ \therefore EB = EF$，$ \angle EAB = \angle EBA = 15^{\circ}$，$ \angle EAF = \angle EFA = 15^{\circ}$。$ \therefore \angle BEC = \angle EAB + \angle EBA = 30^{\circ}$，$ \angle CEF = \angle EAF + \angle EFA = 30^{\circ}$。$ \therefore \angle BEF = 60^{\circ}$。$ \therefore \triangle BEF$ 是等边三角形。