Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

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## Thursday, October 29, 2015

### Geometry Problem 1159: Triangle, Circumcircle, Sagitta, Chord, Arc, Metric Relations

Labels:
chord,
circle,
circumcircle,
metric relations,
sagitta,
triangle

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Join B to O Fröm pythagora theorem r2 = 3/2 sq2R (tr BEO and BDE) The same r3 = 1/2sq10R

ReplyDeleteFrom similar tr BDE and BHF we get BH = 3sq5

http://s29.postimg.org/8hhggawgn/pro_1159.png

ReplyDeleteLet R= radius of circle O

We have ∡ (BHD)= ∡ (BHF)=90 => D, H, F are collinear

DB^2= DẸ 2R = 18.R

BF^2= FG. 2R= 10.R

DB/ BF= sqrt(18/10)= sqrt(1.8)

Observe that F is the midpoint of arc BC so

∡ (CBF)= ∡ (BDF)= angle A/2

So triangle BDH simillar to FBG

DB/BF= BH/FG = sqrt(1.8)

BH= 5 x sqrt(1.8)= 3.sqrt(5)

Tr.s BDE and BHF are similar so

ReplyDeleteBD/BF = 9/BH ....(1)

Tr. s BDH and BGF are similar so

BD/BF = BH/5 ....(2)

From (1) and (2),

9/BH = BH /5

Hence BH = 3sqrt5

Sumith Peiris

Moratuwa

Sri Lanka