7. 如图18-24,在四边形$ABCD$中,$BC = CD$,$∠C = 2∠BAD$,$O是四边形ABCD$内一点,且$OA = OB = OD$.

求证：(1)$∠BOD = ∠C$；
(2)四边形$OBCD$是菱形.

(1)延长AO到点E，∵ OA = OB，∴ ∠ABO = ∠BAO。又∵ ∠BOE = ∠ABO + ∠BAO，∴ ∠BOE = 2∠BAO，同理∠DOE = 2∠DAO，∴ ∠BOE + ∠DOE = 2∠BAO + 2∠DAO = 2(∠BAO + ∠DAO)，即∠BOD = 2∠BAD。又∵ ∠C = 2∠BAD，∴ ∠BOD = ∠C。 (2)连接OC，∵ BC = CD，OB = OD，OC = OC，∴ △OBC ≌ △ODC，∴ ∠BOC = ∠DOC，∠BCO = ∠DCO。∵ ∠BOD = ∠BOC + ∠DOC，∠BCD = ∠BCO + ∠DCO，∴ ∠BOC = $\frac{1}{2}$∠BOD，∠BCO = $\frac{1}{2}$∠BCD。又∵ ∠BOD = ∠BCD，∴ ∠BOC = ∠BCO，∴ OB = BC。又∵ OB = OD，BC = CD，∴ OB = BC = CD = OD，∴ 四边形OBCD是菱形。

1. 如图18-25,$O为正方形ABCD对角线AC$的中点,$\triangle ACE$为等边三角形. 若$AB = 2$,则$OE$的长度为()

A.$\frac{\sqrt{6}}{2}$
B.$\sqrt{6}$
C.$2\sqrt{2}$
D.$2\sqrt{3}$

B

2. 四边形$ABCD$是菱形,添加一个条件：,可使它成为正方形.

答案不唯一，如：∠BAD = 90°

3. 如果一个四边形既是菱形又是矩形,那么它一定是.

正方形

4. 如图18-26,菱形$ABCD的面积为120\ cm^{2}$,正方形$AECF的面积为72\ cm^{2}$,则菱形的边长为(结果中如有根号,请保留根号).

$2\sqrt{34}$

5. 如图18-27,正方形$ABCD$的边长为8,点$E是CD$的中点,$HG垂直平分AE且分别交AE$,$BC于点H$,$G$,则$BG = $______.

1 提示：连接AG，EG。∵ E是CD的中点，∴ DE = CE = 4。设CG = x，则BG = 8 - x。在Rt△ABG和Rt△GCE中，根据勾股定理，得AB² + BG² = AG² = EG² = CE² + CG²，∴ 8² + (8 - x)² = 4² + x²，解得x = 7，∴ BG = BC - CG = 8 - 7 = 1。