Thursday, September 6, 2012

Problem 803: Right Triangle, Altitude, Angle Bisector, Perpendicular, Measurement

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

Click the figure below to see the complete problem 803.

Online Geometry Problem 803: Right Triangle, Altitude, Angle Bisector, Perpendicular, Measurement.

9 comments:

  1. Let AB=c and ∠ (BAD)= ∠ (DAC)= θ
    In 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

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  2. From given picture, it is almost obvious that x=3, so let´s prove it.
    We´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.

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  3. We draw FP⊥AB=>FP=PH=x
    1.ΔBFD=> isosceles => BD=BF

    2.Δ BFP= Δ BDE =>BE=PF(=PH)

    PH=3

    Erina.H-NJ

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  4. Triangles AHF, ABD, DEB are similar.
    X/BD = AH/AB = 3/ BD.
    x = 3

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    Replies
    1. triangle DEB cannot possibly be similar to AHF or ABD

      Delete
  5. Thank you Anonymous:
    First 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

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  6. Triangle AHB ~ triangle BED
    AH/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.

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  7. We'll start with BE=3, FH=x, EF=y, ∠ABH=a, ∠BAD=∠CAD=b.
    By 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

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  8. Problem 803
    Draw 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

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