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There is a square $ABCD$ of side $a$, points $E,F$ lies at centre of respectively $AB,CD$. Line $AE$ intersect with $DF$ at $G$ and $BD$ at $H$. Find area of $DHG$.

I don't know why I can't add a comment but thanks for hint, I have already known how to do it

a busy cat

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Welcome to Math.SE! If this is homework or revision, please say so with a tag. It's best if you show your working so far too :) –  Shaun Nov 18 '13 at 19:11
    
you can check the"check "sign to accept the answer which means your thanks. it will turn green after you check it. –  chenbai Nov 19 '13 at 1:20
    
why do you not see my answers attentively.this is a very nice solution which can help you to solve other hard problems in future –  krishan Nov 21 '13 at 12:32
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5 Answers 5

up vote 1 down vote accepted

Hint: $DHG=DAB-AHB-ADG$.............

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Simpler might be $DHG=DHA-DGA$. –  robjohn Feb 11 at 19:46
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No trig and no long calculations seem to be necessary here.

Drop the perpendiculars from $G$ and $H$ onto $\overline{AD}$ at $I$ and $J$ respectively.

$\hspace{2.5cm}$enter image description here


Since $\triangle JDH\simeq\triangle ADB$, we have that $\overline{JD}=\overline{JH}$

Since $\triangle JAH\simeq\triangle DAE$, we have that $\overline{JA}=2\overline{JH}$

Therefore, $\overline{JH}=\frac13\overline{AD}$


Since $\triangle IDG\simeq\triangle ADF$, we have that $\overline{ID}=2\overline{IG}$

Since $\triangle IAG\simeq\triangle DAE$, we have that $\overline{IA}=2\overline{IG}$

Therefore, $\overline{IG}=\frac14\overline{AD}$


Area of $\triangle DHA=\frac12\overline{JH}\times\overline{AD}=\frac16\overline{AD}^2=\frac16a^2$

Area of $\triangle DGA=\frac12\overline{IG}\times\overline{AD}=\frac18\overline{AD}^2=\frac18a^2$

Therefore, Area of $\triangle DHG=\frac16 a^2-\frac18 a^2=\frac1{24}a^2$

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Would the downvoter care to comment? –  robjohn Feb 13 at 12:50
    
You being a moderator, is it not possible to know who is behind this? –  Sawarnik Feb 13 at 12:59
    
@Sawarnik: ${\small\blacklozenge}$-mods cannot see how anyone votes on a particular post. –  robjohn Feb 13 at 13:11
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@robjohn: shhhhhh... Give away information like that and the proles will start to downvote us with abandon. –  Arthur Fischer Feb 13 at 13:26
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One fact that I have always found useful, is that given two angles and an adjacent side, we can calculate area of a triangle by this simple identity: [Its a good exercise to prove the identity itself, need hints?]

$$Area = \frac{a^2}{2(\cot B + \cot C)}$$

So, in $\triangle DHA$, the identity gives us the area, $\frac{a^2}{2(1 + 2)} = \frac{a^2}{6}$.

And, in $\triangle DGA$, the identity again gives, $\frac{a^2}{2(2 + 2)} = \frac{a^2}{8}$.

Now what would you get when you subtract them both...?

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+1 for the formula, but the geometry is so simple, I don't think trig is really needed. –  robjohn Feb 11 at 20:02
    
@robjohn I saw that that were answers using geometry, so I wanted to give one using simple trig. Actually that formula is helpful in a number of cases, but is not generally introduced, so I wanted to give it here. –  Sawarnik Feb 12 at 8:55
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I see that we were also hit by the downvoter. –  robjohn Feb 13 at 12:49
    
@robjohn Yes, and unlike this time he downvoted it all in one go. –  Sawarnik Feb 13 at 12:53
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NOW Join FC which intersects with DB at point O

  • now see ADEF is a rectengle so G is the surely the midpoint of DF.
  • NOW see CF||AE or in $\bigtriangleup$DFO GH||FO.so H is the midpoint of DO. so we have $\bigtriangleup$DGH=$\frac{1}{4}$.$\bigtriangleup$DFO
  • NOW see that DB=$\sqrt{2}$ a and CF=$\sqrt{a^2+\frac{a^2}{4}}$=$\frac{\sqrt{5}a}{2}$
  • now $\bigtriangleup$DFO IS similar to $\bigtriangleup$DOC as FB||CD
  • so $\frac{DO}{BO}$=2
  • $\frac{DO+BO}{BO}$=3
  • $\frac{DB}{BO}$=3
  • BO=$\frac{\sqrt{2}a}{3}$
  • similarly FO=$\frac{\sqrt{5}a}{6}$
  • now name $\angle$CFB=x and $\angle$DBA=y
  • see tanx=2 and tany=1
  • so now tan$\angle$FOB=tan{180$-$(x+y)}=$-$tan(x+y)=$\frac{tanx+tany}{tanxtany-1}$=3
  • $(sink)^2$+$(cosk)^2$=1 [where k=$\angle$FOB]
  • SO sink=$\frac{3}{\sqrt{10}}$
  • now $\bigtriangleup$FOB=$\frac{1}{2}$.BO.FO.sink=$\frac{a^2}{12}$
  • now $\bigtriangleup$DFB=$\frac{a^2}{4}$
  • so $\bigtriangleup$DFO=$\frac{a^2}{4}$$-$$\frac{a^2}{12}$=$\frac{a^2}{6}$
  • so $\bigtriangleup$DGH=$\frac{1}{4}$$\frac{a^2}{6}$=$\frac{a^2}{24}$enter image description here
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@krishnan If you have used trig, I think my answer is good too :) See it? –  Sawarnik Feb 11 at 16:21
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its a nice solution of course if we know the materials you used,sawarnik. –  krishan Mar 5 at 14:27
    
That formula. You need the proof? –  Sawarnik Mar 5 at 14:34
    
no.i have proved it already. –  krishan Mar 6 at 13:52
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enter image description here

answer

AGH = ABDC - ( ABD + CHD + AGC)

1-□ ABDC

□ ABCD = a*a=a^2..... □ ABCD = (a^2)......(i)

2- △ ABD

△ ABD = (a*a)/ 2 = a^2 /2.... △ ABD =(1/2) a^2......(ii)

3- △ CHD

△ CHD=( HD*KC) / 2....

3-1.Now:KC

△ KCD....KC=KD and CD =a....

KC^2 + KD^2 = CD^2.... <==> 2KC^2 = CD^2 = a^2.... <==> KC^2 = (a^2) / 2........ <==> KC= a/√2....

<==> KC= (a√2)/2......(1)

3-2.Now:HD

FB∥ CF ....F is the midpoint of CD....CF = FD = a/2 <==> CF/ FD = HI/ ID = 1 <==> HI= ID .......(2)

EC ∥ BF....G is the midpoint of AF....AG=GF <==> AG/GF = AH/HI =1 <==> AH= HI .......(3)

from (ii) and (iii) in AD: AH=HI=ID <==> AD= 3 ID) <==> HD =( 2/3)AD....(4)

△ ADB....AD^2 = AB^2 + BD^2 <==> AD^2 = a^2 + a^2 <==> AD = √2a...(5)

from (4) and (5)... HD = (2/3)√2a....(6)

So: △ CHD=( HD*KC) / 2 <==> ( (2/3)√2a)*((a√2)/2)) / 2

<==>△ CHD =( 1/3) a^2....(iii)

4- △ AGC

△ AGC= (AC*LG )/2 = ( a* (LG)) / 2

4-1.Now LG

△ ACF ...AL=LC and AG=GF and LG ∥ CF <==> LG= (1/2) CF = (1/2) (a/2) = a/4

<==> LG= a/4

So:△ AGC= (AC*LG )/2 = ( a* (LG)) / 2 = (a*(a/4)) /2 =(a^2 / 8)

<==> △ AGC = (1/8) a^2....(iiii)

================== Now from AGH = ABDC - ( ABD + CHD+AGC) and (i) (ii) (iii) (iiii)

AGH =a^2 - ( (1/3) a^2 + (1/8) a^2 + (1/2)a^2 )

Finallly area △ AGH = (1/24) a^2


DONE

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what language it is MaDiha –  krishan Nov 24 '13 at 4:45
    
its MaDiha lu lu lu .............. language. –  krishan Feb 13 at 12:33
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