Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

# 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!!!

### 1 Answer

Rating
• Favorite 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

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

Still have questions? Get your answers by asking now.