Proposed Problem
Click the figure below to see the complete problem 384.
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Complete Problem 384
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Thursday, November 12, 2009
Problem 384: Triangle, Angle bisector, Perpendicular, Parallel, Semiperimeter
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Extend AC on either side naming the end to the left of A as X and to the right of C as Y. Draw BM perpendicular to AC. ADBM is concyclic and hence ang. DAX = ang.DBM. But ang. DAX =ang. DAB = ang. DMB. In other words, ang. DBM = ang. DMB or triangle DMB is isosceles which implies that D lies on the perpendicular bisector of BM. Similarly E lies on the perpendicular bisector of BM. In other words DE, the perpendicular bisector of BM is parallel to AC since BM is perpendicular to AC. Further,DE being the perpendicular bisector of BM passes thru. midpoints P and Q rsply. of AB & AC. Now ang. DBA= 90 - (180 -A)/2 = A/2 and ang. BPQ = ang. A =ang. DBA + ang. BDP = A/2 + ang. BDP or ang. BDP=A/2 which makes DP=PB=AB/2 and similarly EQ=BC/2 while PQ=AC/2 since DE is the perpendicular bisector of BM. Thus x = DP+PQ+QE =(AB+BC+CA)/2=(a+b+c)/2=s
ReplyDeleteAjit
Extend ray DA and ray EC to meet at I'.
ReplyDeleteI' is the excentre of triangle ABC w.r.t AC.
Let DE intersect AB at P and BC at Q.
Let XYZ denote angle XYZ if not otherwise stated.
DAB = 90 - A/2 = B/2 + C/2.
=> ABD = A/2.
similarly, BCE = 90 - C/2 = A/2 + B/2.
and CBE = C/2.
now, I'B is the angle bisector of B.
hence, I'BA = I'BC = B/2.
=> I'BD = B/2 + A/2 and I'BE = B/2 + C/2.
now, I'DB + I'EB = 90 + 90 = 180.
=> I'DBE is cyclic.
=> I'BD = I'ED = A/2 + B/2 = BCE.
=> CEQ = QAC. => CQ = QE.
similarly I'DE = I'BE = B/2 + C/2.
=> ADP = DAP. => PD = PA.
In triangle BI'C, I'BC = B/2, BCI' = 90 + C/2 (C + 90 - C/2).
=> EI'B = CI'B = A/2.
=>EDB = B/2 = DBP = BDP.
=> DP = PB.
similarly, AI'B = C/2 = DI'B = DEB = QEB = QBE.
=>QE = QB.
so, P,Q are midpoints of AB and BC.
so DE || AC.
also, DP = AB/2, PQ = AC/2, QE = BC/2.
=> DE = (AB + BC + CA)/2 = s.
Now, another proof :
ReplyDeleteConsider a triangle BAC.
Let P,Q be the midpoints of AB and BC respectively.
Let PQ meet the external angle bisector of A at D.
Let PQ meet the external angle bisector of C at E.
now, BPQ = A = DPA.
PAD = 90 - A/2 => PDA = 90 - A/2.
=> PD = PA. => BDA=90.(since P is the circumcentre of BDA).
similarly, BEC=90.
DE || AC. and DE = s.
https://photos.app.goo.gl/QvHV7HEhZcHco2gw2
ReplyDeleteLet BD and BE meet AC at F and K
Note that triangles BAF and BCK are isosceles
And BD=DF , BE=EK
AF=AB=c and CK=CB= a
So DE//AC and DE= ½ FK= s
Extend BD to meet CA at F and BE to meet AC at G
ReplyDeleteObserve that triangles AFB (AF=AB=c) and CGB (CG=CB=a) are isoscelese and since AD_|_BF => D is the midpoint of BF
similarly E is the mid-point of BG
Hence in the triangle FBG, FG=a+b+c and from mid-point theorem DE is parallel to FG and is half of it
Let BD, CA extended meet at P and let AC, BE extended meet at Q
ReplyDeleteSince Tr.s ADP & ABD are congruent ASA and similarly Tr.s CEQ & BCE, D is the midpoint of BP and E is the BQ
Further AP =mc and CQ = a
Hence applying the midpoint theorem to Tr. PBQ, the result follows
Sumith Peiris
Moratuwa
Sri Lanka
Solution 2
ReplyDeleteLet DE cut AB at X and BC at Y and let DA and EC extended meet at Z
Since ZD & ZC are external bisectors of angles A and C respectively, ZB bisects < ABC
So < DBZ = A/2 + B/2 = < DEZ (ZDBE being concyclic) = 90 - C/2 = < YCE
It follows that YE = YC and hence Y is the centre of Right Triangle BEC
So YE = a/2
Similarly DX = c/2
and XY = b/2 (applying the midpoint theorem to Triangle ABC)
Adding, DE = DX + XY + YE = (a+b+c)/2
Sumith Peiris
Moratuwa
Sri Lanka