Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem.Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 999.
See http://bleaug.free.fr/gogeometry/999.pngBC perpendicular bisector FEG meets AC in F midpoint of AC, and E midpoint of DC (Thales). By construction ∆BEC and ∆BFC are isosceles. Right angle B implies ∆DEB is isosceles and similar to ∆BAE with acute vertex angle = 2x. Let H be the intersection of CD and BF. Since ∆BFC is isosceles, so is ∆BHE. Triangles ∆BAE and ∆BHE are isosceles and share the same base, hence AH bisects angle ∠BAE.∠BAH = ∠BCH implies ∆AHC isosceles. Hence 3x=45° or x=15°.bleaug
Please elaborate on why "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.
as you AHC question why is isosceley
why "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.????????
Draw circle diameter AC, center F. B will be on this circleLet assume that AHC is not isosceles and AB >BCPoint H is on radius FB. M and N are the projection of H over AB and ACUsing elementary geometry we can show that HA > HC and HN> HMHN/HC > HM/HA so ∠(HCB)> ∠(HAB)So If AB> BC , it doesn’t exist point H on FB so that ∠(HCB)= ∠(HAB) ( both angles are positive)
sorry Mr peter ((Using elementary geometry we can show that HA > HC and HN> HMHN/HC > HM/HA so ∠(HCB)> ∠(HAB)))???????????
A solution is here: http://mfcosmos.com/archives/15066
An equally challenging version of the above exercise (with solution) you can find here http://mfcosmos.com/archives/15173
Easily Tr. ABE is isoceles. Let AE = AB = q, DE = EC = BE = p and BD = rNow BE is a tangent to circle ADE hencep^2 = rq ......(1)Further AD is a tangent to circle ACE hence(q-r)^2 = 2p^2 ....(2)From (1) and (2) we have q^2 = 4p^2 -r^2 ......(3)Now consider h the altitude of Tr. ABEh^2 = p^2 -r^2/4 = 4q^2 from .....,(3)So h = q/2 hence 2x = 30 and so x = 15Sumith PeirisMoratuwaSri Lanka
H is the altitude of Tr. ABE drawn from E
Alternatively from (3) above 4p^2 - r^2 = q^2LHS = BC^2 from right Tr. BCDSo BC = q and < BAC = 45 and hence 3x = 45 and x = 15