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6. 如图21-20,在四边形ABCD中,对角线AC与BD相交于点O,OA=OC,OB=OD,且∠AOB=2∠DAO.
图21-20
(1)求证:四边形ABCD是矩形;
(2)若∠AOB : ∠ODC = 4 : 3,求∠ADO的度数.
图21-20
(1)求证:四边形ABCD是矩形;
(2)若∠AOB : ∠ODC = 4 : 3,求∠ADO的度数.
答案
6. (1)$\because OA=OC,OB=OD,\therefore$ 四边形$ABCD$是平行四边形.
$\because ∠ AOB= ∠ DAO + ∠ ADO = 2∠ DAO,\therefore ∠ DAO=∠ ADO. \therefore OA=OD. \therefore AC=BD. \therefore$ 四边形$ABCD$是矩形.
(2)$\because$ 四边形$ABCD$是矩形,$\therefore AB// CD. \therefore ∠ ABO=∠ ODC. \because ∠ AOB: ∠ ODC=4:3,\therefore ∠ AOB: ∠ ABO=4:3. \therefore ∠ BAO: ∠ AOB: ∠ ABO=3:4:3. \therefore ∠ AOB=72°. \therefore ∠ ADO=\dfrac{1}{2}∠ AOB=36°$.
$\because ∠ AOB= ∠ DAO + ∠ ADO = 2∠ DAO,\therefore ∠ DAO=∠ ADO. \therefore OA=OD. \therefore AC=BD. \therefore$ 四边形$ABCD$是矩形.
(2)$\because$ 四边形$ABCD$是矩形,$\therefore AB// CD. \therefore ∠ ABO=∠ ODC. \because ∠ AOB: ∠ ODC=4:3,\therefore ∠ AOB: ∠ ABO=4:3. \therefore ∠ BAO: ∠ AOB: ∠ ABO=3:4:3. \therefore ∠ AOB=72°. \therefore ∠ ADO=\dfrac{1}{2}∠ AOB=36°$.
7. 如图21-21,在$□ ABCD$中,AC,BD相交于点O. E,F分别是OA,OC的中点.
(1)求证:$BE=DF$.
(2)设$\frac{AC}{BD}=k$,当k为何值时,四边形DEBF是矩形?请说明理由.
图21-21
(1)求证:$BE=DF$.
(2)设$\frac{AC}{BD}=k$,当k为何值时,四边形DEBF是矩形?请说明理由.
图21-21
答案
7. (1)如图,连接$DE,BF. \because$ 四边形$ABCD$是平行四边形,$\therefore OB=OD,OA=OC. \because E,F$分别是$OA,OC$的中点,$\therefore OE=\dfrac{1}{2}OA,OF=\dfrac{1}{2}OC. \therefore OE=OF. \because OB=OD,OE=OF,\therefore$ 四边形$BFDE$是平行四边形. $\therefore BE=DF$.
(2)当$k=2$时,四边形$DEBF$是矩形. 理由如下:当$BD=EF$时,四边形$DEBF$是矩形,$\therefore$ 当$OD=OE$时,四边形$DEBF$是矩形. $\because AE=OE,\therefore AC=2BD. \therefore$ 当$k=2$时,四边形$DEBF$是矩形.
