# My son problem

#### marcus2

Hi there,

My son ask me to help him with one math question which seems to be very easy (as this is primary school) but I cannot help him because I finished my school about 30 years ago and do not remember. So your help needed. The question is:

Given is rectangle ABCD and points E and F that lie on sides BC and CD, where <EAF=45 degrees and BE=DF. Proof that field of triangle AEF is the same as sum of triangle ABE and ADF.

Marek

#### Attachments

• 72.4 KB Views: 10

#### tahirimanov19

Let sides of the rectangle be $a, b$. And $BE=DF=c$.

Area of triangle $ABE = ac/2$ and area of triangle $ADF = bc/2$. And the area of $CFE = (a-c)(b-c)/2=(ab-ac-bc+c^2)/2$.

And the area of the rectangle is $ab$.

Therefore, the area of the triangle $AEF = ab- \dfrac{ac}{2} - \dfrac{bc}{2} - \dfrac{ab-ac-bc+c^2}{2} = \dfrac{ab-c^2}{2}$.

Which means they are not always equal. (Probably never equal.)

Forum Staff
They're equal.

#### DarnItJimImAnEngineer

tahirimanov, you're claiming the areas will not be equal because random choices of a, b, and c don't satisfy the equation, right?

Except you're forgetting the 45Â° constraint. There are only two degrees of freedom in defining the rectangle/triangle pair -- we can choose a and b, or a and c, or a and <DAF, etc..

I have confirmed the areas are equal. However, I haven't been able to come up with a proof that doesn't involve the law of sines or other trig identities. Since it was stated to be a primary school problem, there is probably a much simpler but not obvious way to show this. Perhaps somebody needs to print out an example and start cutting and rearranging paper for inspiration.

• 2 people

#### tahirimanov19

I take my last statement back. I made an error on simulation and quickly jumped to the conclusion.
Returning to the solution:
Let sides of the rectangle be $a, b$. And $BE=DF=c$.

Area of triangle $ABE = ac/2$ and area of triangle $ADF = bc/2$. And the area of $CFE = (a-c)(b-c)/2=(ab-ac-bc+c^2)/2$.

And the area of the rectangle is $ab$.

Therefore, the area of the triangle $AEF = ab- \dfrac{ac}{2} - \dfrac{bc}{2} - \dfrac{ab-ac-bc+c^2}{2} = \dfrac{ab-c^2}{2}$.

Draw a perpendicular from F to AB and let intersection be Q, $AQ=c$. The intersection of AE and FQ is R. $QR = c^2/a$.
Therefore $FR=b-\dfrac{c^2}{a}$

Can't find a way without invoking trigonometry.

#### marcus2

Thank you tahirimanov19, but as I understand gives us only that AEF = (ab-c^2)/2 and if FR=b-c^2/a means that AEF = a*(b-c^2/a)/2 which mean a*FR/2. Do not still see how to compare to sum of fields of triangles AFD and ABE is equal to AEF means ac/2+bc/2=a*FR/2. On attached picture I measured b=9 and a=14, c=4,5 and FR=7,5. There are some roundings issues but sum of triangles ac/2+bc/2= 52 (ca) and a*FR/2=52,5 (so rounding to 52 due to some inaccuracy of my measurement). Means this is equal but still no proof. I also measure that point E split b into equal parts of c. So maybe trigonometry is needed (or properties of triangles) to proof - I tried with sin45*AE*AF but led me to nothing.

#### Attachments

• 75.2 KB Views: 3

#### skipjack

Forum Staff
It's useful to consider a height for triangle AEF, which could easily be done using trigonometry, but also, given the known angle, can be done by geometry (Pythagoras). Also apply Pyhtagoras to triangles ABE and ADF. Now apply a little algebra to get the desired result.