Online Geometry theorems, problems, solutions, and related topics.
Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 789.
http://img138.imageshack.us/img138/1382/problem789.png( complete solution)drawing lines and circle per attached sketchWe 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) = ∠ABDSo ABD is an isosceles triangle with AB=ADWe also have ∠HBD=∠DBK=∠BDK.. ( alternate angles)So BKD is an isosceles triangle with KB=KDA,I and K are collinear ....(located on perpen. Bisector of BD) IK is an angle bisector of ∠ BKDSo IM=IL=ED=r
Anonymous said... AB=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
Simply let AB = a, BC = b, AC = cBy using the properties of inradius, r = ab/(a+b+c)Our target is DE = ab/(a+b+c)The altittude = ab/cCH = sqrt(b^2-(ab/c)^2) = (b^2)/cUsing angle bisector theorem, CD = CH*(BC/(BC+CH)) = (b^2)/(a+c)By incircle properties, CE = (b+c-a)/2Then DE = CE - CD = (b+c-a)/2 - (b^2)/(a+c) By trigonometry, we can easily prove the remainings.
Anonymous1. AB=AD We put M and N the points where they match with AB and BC IM=EN=NB=BM=rAM=AE AD-AE= AB-AM=r so in this case ED=r
http://img507.imageshack.us/img507/9789/p781resuelto.png(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=rGreetings :)
Look at THIS picture! Sorry :Phttp://img27.imageshack.us/img27/8784/p789resuelto.png
Let F be the tangency point on AB< IBC = 45 and < DBC = 45-C/2 so < IBD = C/2Hence BCDI is cyclic and BI = ID since IC bisects < CSo Tr.s BFI and DEI are congruent (SS & right angle) Therefore DE = BF = rSumith PeirisMoratuwaSri Lanka
Or because BCDI is cyclic and < IDE = IBC = 45 the result follows
Problem 789Let <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