Find the area of triangle 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: 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.
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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 $\triangle DHG=\frac16 a^2-\frac18 a^2=\frac1{24}a^2$
A: Hint:  $\text{area of }\triangle DHG=\text{area of }\triangle DAB-\text{area of }\triangle AHB-\text{area of }\triangle ADG$.............
A: 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...?  
A: 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}$
A: 
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=( HDKC) / 2
<==> ( (2/3)√2a)((a√2)/2)) / 2
<==>△ CHD =( 1/3) a^2....(iii)
4- △ AGC
△ AGC= (ACLG )/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= (ACLG )/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
