## Saturday, March 29, 2014

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

Geometry Problem.

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

Click the figure below to see the complete problem 999.

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

BC 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

1. Bleaug

Excellent solution.

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

1. as you AHC question why is isosceley

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

3. Draw circle diameter AC, center F. B will be on this circle
Let 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)

4. sorry Mr peter ((Using elementary geometry we can show that HA > HC and HN> HM
HN/HC > HM/HA so ∠(HCB)> ∠(HAB)))???????????

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

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

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

Now 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

6. H is the altitude of Tr. ABE drawn from E

7. Alternatively from (3) above

4p^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