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two simple geometry questions
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
2. In the figure below, OF=2FC. Find the value of x
thank you very much!!!
- ☂雨後晴空☀Lv 71 decade agoFavorite Answer
1)ㄥPAO = ㄥRBO = 90 (given)
ㄥPOA = ㄥROB(對頂角)
Hence △PAO ~ △RBO (A.A.A.)
PO / AO = OR / OB
AO * OR = PO * OB
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