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

Figure of the problem

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.

  • $\begingroup$ could you please double check if you've named the points properly in your answer as $\triangle ABC$ does not exist and it is not equivalent to what you've written. $\endgroup$ – The Integrator Apr 9 '18 at 17:36
  • $\begingroup$ @PranavB23 I corrected my edit. Sorry for the delay. Went for dinner. $\endgroup$ – Saradamani Apr 9 '18 at 17:55

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

  • $\begingroup$ Your answer is also beautiful. These solutions are like the pearls of a necklace. I am in love with these.So many wonderful ideas!! Just looked at the problem from another perspective!! $\endgroup$ – Saradamani Apr 10 '18 at 4:06

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?

  • $\begingroup$ the best answer! $\endgroup$ – Orest Bucicovschi Apr 9 '18 at 21:49

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 $\displaystyle \frac{1}{2}(9)(6)=27$.

  • $\begingroup$ Can you please explain and elucidate The distance from $F$ to $DE$ is $\displaystyle 18\times \frac{1}{3}=6$. The area of $\triangle DEF$ is $\displaystyle \frac{1}{2}(9)(6)=27$. $\endgroup$ – Saradamani Apr 9 '18 at 18:04
  • $\begingroup$ No need for elaboration. Got your answer. Thanks .. $\endgroup$ – Saradamani Apr 9 '18 at 18:14
  • $\begingroup$ Superb observation. This seems I have not been in touch with Geometry for long. Had I been in high school now, I could have solved it anyway. $\endgroup$ – Saradamani Apr 9 '18 at 18:15

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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 $k = d$.



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