Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to see the complete problem 803.
Thursday, September 6, 2012
Problem 803: Right Triangle, Altitude, Angle Bisector, Perpendicular, Measurement
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Let AB=c and ∠ (BAD)= ∠ (DAC)= θ
ReplyDeleteIn right triangles ABD and BED we have BD=c. tan(θ) and BE=BD.cos(2 θ)= c.tan(θ).cos(2 θ)
In right triangles ABH and AFH we have AH=c.cos(2 θ) and FH=AH.tan(θ)=c.tan(θ).cos(2 θ)=3
From given picture, it is almost obvious that x=3, so let´s prove it.
ReplyDeleteWe´ll start with same denoting: AB = c, BC = a, AC = b, AH = p, BH = h, BE = z.
Note that we´ll use twice the angle bisector theorem, which says that angle bisector divides the opposite side of a triangle in the same proportion as the proportion of adjoin sides. In other words, theorem used to triangle AHB says:
c/p = (h-x)/x, from which x = hp/(c+p) immediately follows.
We use our theorem once more on triangle ABC:
c/b = BD/CD = z/(h-z), the second equality is given by similarity of triangles BED and BHC. So we have c/b = z/(h-z), thus z = hc/(b+c).
Now, let´s divide these expressions of x and z:
x/z = (hp/(c+p)) / (hc/(b+c)) = (p(b+c)) / (c(c+p)). Note that p can be easily replaced by c^2/b form Euclidean Theorem for side AB. After replacing p, it´s straighforward to find out that x/z = 1, thus x=z and finally, using z=3, we can see that x=3 as well.
We draw FP⊥AB=>FP=PH=x
ReplyDelete1.ΔBFD=> isosceles => BD=BF
2.Δ BFP= Δ BDE =>BE=PF(=PH)
PH=3
Erina.H-NJ
Triangles AHF, ABD, DEB are similar.
ReplyDeleteX/BD = AH/AB = 3/ BD.
x = 3
triangle DEB cannot possibly be similar to AHF or ABD
DeleteThank you Anonymous:
ReplyDeleteFirst line should read "∆s AHF,ABD are ∼ ; ∆s ABH,BDE are ∼ "
Details:
∠FAH = ∠BAD, ∠AHF = 90° = ∠ABD imply ∆AHF ∼ ∆ABD.
∴ x/BD= AH/AB.
∠ABH = ∠BCH = ∠EDB (corresp angles) and ∠AHB = 90° = ∠BED imply ∆ABH ∼ ∆BDE.
∴ AH/BE = AB/BD and so AH/AB = BE/BD
Hence x/BD = BE/BD, x = BE = 3
Triangle AHB ~ triangle BED
ReplyDeleteAH/AB=BE/BD…csct
AH/AB=3/BD….(BE=3)…(1)
Triangle ABD ~ triangle AHF
AH/AB=HF/BD….csct
AH/AB=x/BD…(2)
By (1) and (2),
3/BD= AH/AB=x/BD
3/BD =x/BD
3=x.
We'll start with BE=3, FH=x, EF=y, ∠ABH=a, ∠BAD=∠CAD=b.
ReplyDeleteBy the angle bisector theorem, BA:AH=BF:FH=3+y:x
By the exterior angle theorem for ΔABF, ∠BFD=a+b
∠ACB=∠ABH=a
By the exterior angle theorem for ΔACD, ∠BDF=a+b
∴BF=BD=3+y
For ΔABH and ΔBDE, ∠BAH=∠DBE=2b, ∠ABH=∠BDE=a,
So, ΔABH is similar to ΔBDE
∴BA:AH=DB:BE=3+y:x
BE=FH=x=3
Problem 803
ReplyDeleteDraw DP perpendicular in AC ( P owned the segment AC.Is BD=DP=EH.But <BFD=<FAB+<FBA=<DAC+<DCA=<BDF then BF=BD=EH or BE+EF=EF+FH.So
X=FH=BE=3.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE
Draw a perpendicular DG from D to AC.
ReplyDeleteTr.s ABD and AGD are congruent ASA
BF = BD
DEHG is a rectangle
So EH = DG =BD = BE
So EF + x = EF + 3
And so x = 3
Sumith Peiris
Moratuwa
Sri Lanka
Denote the foot of the perpendicular from D to AC as P
ReplyDeleteSince m(ABD)=m(APD)=90=> ABDP are concyclic
=>m(DBP)=m(DPB)=A/2
=>BD=DP --------(1)
Observe that triangle BFD is isosceles with BF=BD
=>BF=DP
=>x=3