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

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

Click the figure below to enlarge the figure of problem 1011.

## Sunday, May 4, 2014

### Geometry Problem 1011: Triangle, Angle Bisector, Vertex, Circle, Center, Radius, Parallel Lines

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Let BC=a, AC=b, AB=c.

ReplyDeleteThen

AE = bc/(a+b)

EB = ac/(a+b)

AD = bc/(a+c)

DC = ab/(a+c)

AG = a+c

AF = a+b

Thus

AE/AG = AD/AF = bc/[(a+b)(a+c)]

Hence, DE//FG.

Below is geometry solution.

ReplyDeleteNote that ∠ (CBG)=1/2.( 180-∠A-∠C)=1/2. ∠B => DB//CG

Similarly EC//BF

∆AEC similar to ∆ABF => AE=(AB.AC)/ AF…… (1)

∆ADB similar to ∆ACG => AD=(AB.AC)/ AG….. (2)

Divide ( 1) to (2) we get AE/AG=AD/AF => ED//FG

Problem 1011

ReplyDeleteIs <BCA=<CFB+<CBF or 2.<BCE=2.<BFC or <BCE-<BFC so CE//BF. Then AC/AE=AF/AB (1).

But <CBA=<BCG+<BGC or 2.<DBA=2.<CGB so DB//GC.Then AG/AC=AB/AD (2). With multiplication the (1) and (2) we have (AC/AE).(AG/AC)=(AF/AB).(AB/AD) or AG/AE=AF/AD. So DE//FG.

APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE

Let BC = CF = BG = a, AE =r, EB = s, AP = p and DC = q.

ReplyDelete(r+s)/a = p/q ….(1) and

(p+q)/a = r/s ….(2)

From (1); (r+s+a)/a = (p+q)/q and

from (2); (p+q+a)/a = (r+s)/s and upon dividing

(r+s+a)/(p+q+a) = (p+q)/q X s/(r+s) = {(p+q)s}/{q(r+s)} = {ar}/{ap} =r/p

So AG/AF = AE/AD and hence DE // GF

Sumith Peiris

Moratuwa

Sri Lanka