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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 803.
Let AB=c and ∠ (BAD)= ∠ (DAC)= θIn 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.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.
We draw FP⊥AB=>FP=PH=x1.ΔBFD=> isosceles => BD=BF2.Δ BFP= Δ BDE =>BE=PF(=PH)PH=3Erina.H-NJ
Triangles AHF, ABD, DEB are similar.X/BD = AH/AB = 3/ BD.x = 3
triangle DEB cannot possibly be similar to AHF or ABD
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/BDHence x/BD = BE/BD, x = BE = 3
Triangle AHB ~ triangle BEDAH/AB=BE/BD…csctAH/AB=3/BD….(BE=3)…(1)Triangle ABD ~ triangle AHFAH/AB=HF/BD….csctAH/AB=x/BD…(2)By (1) and (2),3/BD= AH/AB=x/BD3/BD =x/BD3=x.
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:xBy the exterior angle theorem for ΔABF, ∠BFD=a+b∠ACB=∠ABH=aBy the exterior angle theorem for ΔACD, ∠BDF=a+b∴BF=BD=3+yFor ΔABH and ΔBDE, ∠BAH=∠DBE=2b, ∠ABH=∠BDE=a,So, ΔABH is similar to ΔBDE∴BA:AH=DB:BE=3+y:xBE=FH=x=3
Problem 803Draw 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.SoX=FH=BE=3.APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE