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.

## Saturday, March 29, 2014

### Geometry Problem 999. Right Triangle, Midpoint, Median, Double Angle, Congruence

Labels:
congruence,
double angle,
midpoint,
right triangle

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See http://bleaug.free.fr/gogeometry/999.png

ReplyDeleteBC 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

Bleaug

DeleteExcellent solution.

Please elaborate on why "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.

ReplyDeleteas you AHC question why is isosceley

Deletewhy "∠BAH = ∠BCH implies ∆AHC isosceles." Thanks.????????

DeleteDraw circle diameter AC, center F. B will be on this circle

DeleteLet assume that AHC is not isosceles and AB >BC

Point H is on radius FB. M and N are the projection of H over AB and AC

Using elementary geometry we can show that HA > HC and HN> HM

HN/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> HM

DeleteHN/HC > HM/HA so ∠(HCB)> ∠(HAB)))???????????

A solution is here: http://mfcosmos.com/archives/15066

ReplyDeleteAn equally challenging version of the above exercise (with solution) you can find here http://mfcosmos.com/archives/15173

ReplyDeleteEasily Tr. ABE is isoceles. Let AE = AB = q, DE = EC = BE = p and BD = r

ReplyDeleteNow BE is a tangent to circle ADE hence

p^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. ABE

h^2 = p^2 -r^2/4 = 4q^2 from .....,(3)

So h = q/2 hence 2x = 30 and so x = 15

Sumith Peiris

Moratuwa

Sri Lanka

H is the altitude of Tr. ABE drawn from E

ReplyDeleteAlternatively from (3) above

ReplyDelete4p^2 - r^2 = q^2

LHS = BC^2 from right Tr. BCD

So BC = q and < BAC = 45 and hence 3x = 45 and x = 15