Geometry Problem

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

Click the figure below to see the complete problem 789.

## Monday, July 23, 2012

### Problem 789: Right Triangle, Altitude, Angle Bisector, Incircle, Tangency Point, Inradius, Circles, Tangent Line

Labels:
altitude,
angle bisector,
incircle,
inradius,
right triangle,
tangency point,
tangent

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http://img138.imageshack.us/img138/1382/problem789.png

ReplyDelete( complete solution)

drawing lines and circle per attached sketch

We have Arc B’F= Arc FC and Arc AB=Arc AB’

∠ABD=1/2(Arc AB’ +Arc B’F)

∠ADB=1/2(Arc AB +Arc FC) = ∠ABD

So ABD is an isosceles triangle with AB=AD

We also have ∠HBD=∠DBK=∠BDK.. ( alternate angles)

So BKD is an isosceles triangle with KB=KD

A,I and K are collinear ....(located on perpen. Bisector of BD)

IK is an angle bisector of ∠ BKD

So IM=IL=ED=r

Anonymous said...

DeleteAB=AD so we put M and N the match points with AB and BC .

IM=IN=BN=BM=r AM=AE---> AD-AE = AB-AM=r

ED=r

July 23, 2012 7:35 PM

CD/HD = BC/BH or CD = CH *BC/(BC+BH) = a^2/(b+c). Now let the in-circle touch AB at F. AE =AF = c - r or CE = b-c+r which gives us:DE = CE-CD=b-c+r-a^2/(b+c)= (b^2-c^2-a^2)/(b+c) + r = r since b^2=a^2+c^2. Thus, DE = r

ReplyDeleteSimply let AB = a, BC = b, AC = c

ReplyDeleteBy using the properties of inradius, r = ab/(a+b+c)

Our target is DE = ab/(a+b+c)

The altittude = ab/c

CH = sqrt(b^2-(ab/c)^2) = (b^2)/c

Using angle bisector theorem, CD = CH*(BC/(BC+CH)) = (b^2)/(a+c)

By incircle properties, CE = (b+c-a)/2

Then DE = CE - CD = (b+c-a)/2 - (b^2)/(a+c)

By trigonometry, we can easily prove the remainings.

Anonymous

ReplyDelete1. AB=AD We put M and N the points where they match with AB and BC

IM=EN=NB=BM=r

AM=AE AD-AE= AB-AM=r so in this case ED=r

http://img507.imageshack.us/img507/9789/p781resuelto.png

ReplyDelete(look at the picture)

Let be P the intersection point of the ray AI and segment BC. Note that as triangle ABC is rectangle, we know that <BAC=<HBC, but AI and BD are bisectors, then <DAP=<DBP, i.e, the quadrilateral ADPB is cyclic. immediately <PDA=90° and BP=DP. As triangle DBP is isosceles, the incircle is tangent to DP too.

Now let Q be the tangency point of incircle with DP. Note that IQED is a square of side r. Therefore DE=r

Greetings :)

Look at THIS picture! Sorry :P

Deletehttp://img27.imageshack.us/img27/8784/p789resuelto.png

Let F be the tangency point on AB

ReplyDelete< IBC = 45 and < DBC = 45-C/2 so < IBD = C/2

Hence BCDI is cyclic and BI = ID since IC bisects < C

So Tr.s BFI and DEI are congruent (SS & right angle)

Therefore DE = BF = r

Sumith Peiris

Moratuwa

Sri Lanka

Or because BCDI is cyclic and < IDE = IBC = 45 the result follows

ReplyDeleteProblem 789

ReplyDeleteLet <ABH=<BCA=x and <HBD=<DBC=y then <ABD=x+y=<ADB so AB=AD.Si the incircle of radius r is tangent to AB at F then AF=AE,BF=ED.But <IBF=45 so ED=FB=IF=r.

APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE