【答案】
分析:(1)由BD垂直于AC,得到三角形ABD為直角三角形,根據(jù)AB及sinA的值,利用銳角三角函數(shù)定義求出BD及AD的長(zhǎng),再由AC-AD求出DC的長(zhǎng),在直角三角形BDC中,利用勾股定理即可求出BC的長(zhǎng);
(2)還存在2條其它平分△ABC的周長(zhǎng)和面積的直線,理由為:若直線經(jīng)過(guò)B(或C)點(diǎn),由直線平分△ABC的面積,則直線必經(jīng)過(guò)AC(或AB)的中點(diǎn),而此時(shí)直線必不平分△ABC的周長(zhǎng),故直線不經(jīng)過(guò)△ABC的頂點(diǎn),分兩種情況考慮:(i)直線與AB(或AC)、BC相交,設(shè)直線與AB、BC相交于點(diǎn)D、E,過(guò)A、D分別作BC的垂線,垂足為F、H點(diǎn),如備用圖1所示,假設(shè)DE平分三角形的周長(zhǎng),設(shè)BD=5k,則DF=4k,BE=8-5k,利用三角形的面積公式表示出BDF的面積,根據(jù)此三角形面積為三角形ABC面積的一半列出關(guān)于k的方程,求出方程的解得到k的值,可得出在BC上取BE=5,在BA上取BD=3,過(guò)D、E的直線就是所求的,同理AC,BC相交的直線也存在一條;(ii)直線與AB,AC相交,設(shè)直線與AB,AC分別交于D,E,過(guò)D作DF⊥AC,垂足為F點(diǎn),如備用圖2所示,設(shè)AE=x,則AD=8-x,根據(jù)三角形ADE的面積為三角形ABC面積的一半列出關(guān)于x的方程,求出方程的解得到x的值,經(jīng)判斷不合題意,舍去,綜上,得到滿足題意的直線有2條.
解答:解:(1)∵BD⊥AC,
∴∠ADB=90°,
∵在Rt△ABD中,AB=AC=5,sinA=

,
∴BD=ABsinA=5×

=

,
∴根據(jù)勾股定理得:AD=

=

,
∴DC=AC-AD=5-

=

,
在Rt△BCD中,根據(jù)勾股定理得:BC=

=6;
(2)還存在2條其它平分△ABC的周長(zhǎng)和面積的直線,理由為:
若直線經(jīng)過(guò)B(或C)點(diǎn),由直線平分△ABC的面積,則直線必經(jīng)過(guò)AC(或AB)的中點(diǎn),
而此時(shí)直線必不平分△ABC的周長(zhǎng),故直線不經(jīng)過(guò)△ABC的頂點(diǎn),
分兩種情況考慮:
(i)直線與AB(或AC)、BC相交,設(shè)直線與AB、BC相交于點(diǎn)D、E,過(guò)A、D分別作BC的垂線,垂足為F、H點(diǎn),
如備用圖1所示:

∵AB=AC=5,AH⊥BC,
∴BH=CH=

BC=3,
在Rt△ABH中,根據(jù)勾股定理得:AH=

=4,
∴∠DFB=∠AHB=90°,又∠B=∠B,
∴△BDF∽△BAH,
∴BF:FD:BD=BH:AH:AB=3:4:5,
又∵三角形ABC的周長(zhǎng)為5+5+6=16,
∴BD+BE=8,
設(shè)BD=5k,則DF=4k,BE=8-5k,
∴S
△BDE=

S
△ABC=

BC•AH=6,即

BE•DF=

=6,
整理得:5k
2-8k+3=0,
解得:k=

或k=1(舍去),
這時(shí)在BC上取BE=5,在BA上取BD=3,過(guò)D、E的直線就是所求的,
同理AC,BC相交的直線也存在一條;
(ii)直線與AB,AC相交,設(shè)直線與AB,AC分別交于D,E,過(guò)D作DF⊥AC,垂足為F點(diǎn),
如備用圖2所示:
設(shè)AE=x,則AD=8-x,
∵在Rt△ADF中,sinA=

,

∴DF=ADsinA=

(8-x),
當(dāng)S
△AED=

AE•DF=

•x•

(8-x)=6,
整理得:2x
2-16x+25=0,
解得:x
1=4+

>5(舍去),x
2=4-

,
則AD=8-x=4+

>5(不合題意,舍去),
綜上,還存在2條其它平分△ABC的周長(zhǎng)和面積的直線.
點(diǎn)評(píng):此題考查了相似性綜合題,涉及的知識(shí)有:勾股定理,銳角三角函數(shù)定義,一元二次方程的應(yīng)用,以及解直角三角形,利用了數(shù)形結(jié)合及分類討論的思想,是一道多知識(shí)的探究題.