9.a
BE=ED(given)
EC=EC(common side)
<ECB=180-<ECD
<ECB=90
so,triangleECB=triangleECD(R.H.S)
As,triangleECB=triangleECD(R.H.S)
So,BC+CD=BD
As,BC=CD
So,BC=BD-CD
BC=1/2BD
唔知可唔可以咁証明- -"
b.As,BC=CD
BC=BD/2
So,CD=BD/2
AD=AE+ED
AE=ED
ED=AD/2
<EDC=<ADB(common angle)
CD/EB=BD/2/AD/2
=BD/AD
So,triangleABD~triangleECD(ratio of 2 side inc.<)
7a
GQSH is a straight line.(given)
<GQP+<PQS=180(adj. <s on st. line)
<PQS=90
<PQS+RSQ=180
So,AB//CD(int.<s supp.)
b.
As,AB//CD
3x-12=5x/2(alt.<s,AB//CD)
6x-24=5x
x=24
[ 本帖最後由 `乜囧. 於 2009-1-4 01:01 AM 編輯 ]
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