two simple geometry questions

用戶11165

2010-01-22 1:11 am

i just want to verify whether i did them right or not

1. In the figure below, AR and PB are the altitudes of triangle PQR. Prove the product of AO and OR equals the product of PO and OB

圖片參考：http://imgcld.yimg.com/8/n/HA00008075/o/701001210100513873384230.jpg

2. In the figure below, OF=2FC. Find the value of x

圖片參考：http://imgcld.yimg.com/8/n/HA00008075/o/701001210100513873384231.jpg

thank you very much!!!

回答 (1)

用戶2053

2010-01-22 3:04 am

✔ 最佳答案

1)ㄥPAO = ㄥRBO = 90 (given)
ㄥPOA = ㄥROB(對頂角)
Hence △PAO ~ △RBO (A.A.A.)

PO / AO = OR / OB

AO * OR = PO * OB

Q.E.D.

2)Assume that O is the centre.
Joint AC , then ㄥCED = x = ㄥCAD (with common chord DC)
So ㄥBAC = (x+10) - ㄥCAD = (x+10) - x = 10 ,
hence ㄥBOC = 2ㄥBAC = 20
Let OB = OC = 3 , by given that OF = 2FC we get OF = 2 and FC = 1.
In △OBF , by cosine formula :
BF^2 = OB^2 + OF^2 - 2(OB)(OF)cosㄥBOF
BF^2 = 3^2 + 2^2 - 2(3)(2)cos20
BF = 1.3128932
By sine formula :
sinㄥBOF / BF = sinㄥBFO / OB
sin20 / 1.3128932 = sinㄥBFO / 3
sinㄥBFO = 0.7815262
ㄥBFO = 51.4(rejected) or 180 - 51.4 = 128.6
And ㄥBFO = ㄥCED + ㄥEDF
= ㄥCED + ㄥEDO + ㄥODB
= x + x + (180 - 90 - (x+10))
= x + 80
x + 80 = 128.6
x = 48.6

👍 0 👎 0