Area of triangle Problem where two triangles overlap

Question

Let AC and CE be perpendicular line segments,each of length 18.Suppose B and D are the mid points of AC and CE respectively. If F is the intersection of EB and AD ,then the area of triangle DEF is

My attempt is Area of $\Delta \space DEF=($ Area of $\Delta \space ACD$+ Area of $\Delta \space BCE$ - Area of quadrilateral BCDF)/2

as Triangles ABF and DEF are congruent.

Now Quadrilateral BCDEF can be divided by line segment CF into $\Delta \space BCF \space and \space \Delta CFD$

Now $\angle ADC = tan^{-1}(9/18) \implies tan^{-1}(\frac{1}{2}) $

Similarly we can also get $\angle CBF$ and as $\Delta BCF$ is congruent to $\Delta DCF$ , $\angle BFC \space = \angle DFC$ and as $\angle BCF \space = \angle DCF= 45^{\circ}$ , we can get $\angle BFC \space and \space \angle DFC$

Finally using the sine rule we can get BF,CF,FD,CD using which we can calculate the area of the quadrilateral BCDF as a sum of triangles BFC and DFC

We can easily get the area of triangles ACD and BCE as they are right angled triangles and subtracting it by the new found area of quadrilateral BCDF, we can get our result.

I do not want this process. I want an easier approach to this problem. Please suggest with reasons.

Answer 1

Let $[ACE]=S$ denote the area of $\triangle ACE$. The point $F$ is the centroid of $\triangle ACE$, hence \begin{align} |FG|&=\tfrac13|CG|\text{ (CH should be $CG$ , error corrected)} .\\\\ [AFE]&=\tfrac13[ACE]=\tfrac13S,\text{ ($\Delta ACG$ should be $\Delta ACE$, error corrected).} \\\\ [EFC]&=[ACF]=\tfrac12([ACE]-[AFE])=\tfrac12(S-\tfrac13S)=\tfrac13S .\\\\ [DEF]&=[DFC]=\tfrac12[EFC]=\tfrac16S=\tfrac16\cdot\tfrac12\cdot18^2 =27 . \end{align}

Answer 2

Let $G$ be the mid-point of $BC$. By the mid-point theorem, $BE\parallel GD$.

So, by the intercept theorem, $AF:FD=AB:BG=2:1$.

The distance from $F$ to $DE$ is $\displaystyle 18\times \frac{1}{3}=6$.

The area of $\triangle DEF$ is $\displaystyle \frac{1}{2}(9)(6)=27$.

Answer 3

Join the points $C$ and $F$

The area of triangle $CFD$ is the same as the area of triangle $FED$ since they have the same base and height. Also by symmetry, area $BFC$ is equal to each of these.

Can you finish?

Answer 4

When I see similarity I try to noodle about and see if I can break the similarity further.

Area of $DEF$ is obviously the area of $CBE$ minus Area of quadrangle that is the overlap of the two triangles. Area of $CBE = \frac 12*9*18 = 81$. Can I break down the quadrangle? Particularly by looking for similar triangles?

this may not be the slickest idea but it is the one that jumps at me. Make a point, $W$ on $CA$ and a point $V$ on $CE$ so that $CWFV$ is a rectangle. Then triangles $WBF$ and $VDF$ and $CDA$ and $CBE$ are all similar.

So $WB = k*CB = k*9$ and $CW = VF = k*CA = k*18$ for some scaling factor $k$.

So $9= CB = BW + WC = 9*k + 18*k$ so $k = \frac 13$ and $WB= 3$ and $CW = 6$.

So the the quadrangle is the area of the two small triangles ($\frac 12*3*6 = 9$). And the area of the square is $6^2= 36$.

So the area of the quadrangle is $36 + 2*9 = 54$. The area of $CBF = 81$ and so area of $DEF = 81 - 54 = 27$.

Oh.... I took it for granted that $WBF$ and $VDF$ where congruent and not just similar. It's pretty obvious by symmetry and .... Well $9=CD = CV + VD = k*18 + d*9$ (where $k$ is scaling factor of $WBF$ and $d$ is scaling factor of $VFD$) and $9=CB = CW+WB = d*18 + k*9$ so $18k + 9d = 18b + 9k$ so $k = d$.

And

Answer 5

$BG$ is parallel to $CE$ (midline segment).

$F$ is the center of gravity of the triangle, therefore situated at the $2$ thirds of median $CG$ giving (using intercept theorem) $6$ for the height of triangle DEF.

Conclusion : the area of triangle $DEF$ is : $\dfrac12$ base $\times$ height = $\dfrac12 \times 9 \times 6 \ = 27$