## Saturday, September 6, 2014

### Geometry Problem 1042: Scalene Triangle, Isosceles, Equal Angles, Parallelogram, Congruence

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

Click the figure below to view the complete problem 1042. 1. Let BC=k BE, then AB=k AD and AC=k AF.

Let S(P, k, α) be the homothetic rotation transformation,
with center P, scale factor k, and angle α.

Consider first S(B, k, −α) and then S(A, 1/k, α).
B→B→D
E→C→F

On the other hand, the combined transformation
S(A, 1/k, α)。S(B, k, −α) = T(v)
which is a translation by vector v.
[It is because (k)(1/k)=1 and (α)+(−α)=0]

Hence, vector v = vector BD = vector EF.
i.e. BEFD is a parallelogram.

2. Let DF cut BC at M and EF cut AB at N.
Note that ABC is the image of ADF in the spiral similarity transformation with
Rotation angle = α => ∠(DMB)= α => DF//BE
In the similar way we also get EF//BD so DFEB is a parallelogram

3. Tr.s AFC and ADB are similar hence AF / b = AD / c = p say.

So AF = pb and AD = pc and since < DAF = < BAC it follows that Tr.s ADF and ABC are similar the respective sides being in the ratio of p

So DF = pa = BE since tr.s AFC and BEC are similar.

Similarly EF = BD = pc since Tr.s AFC and BEC are similar and Tr.s BEC and ABC are similar

Hence in quadrilateral BDFE the opposite sides are equal and hence BDFE is a parallelogram

Sumith Peiris
Moratuwa
Sri Lanka

4. Problem 1042