Question regarding triangles?

Question

ABC and BDE are two equilateral triangles such that D is the midpoint of BC. A and E lie on opposite sides of BC. Join EC and AD. If AE intersects BD at F, Show that:
(a) ar(BDE) = 1/4 ar(ABC)
(b) ar(BDE) = 1/2 ar(BAE)
(c) ar(ABC) = 2*ar(BEC)
(d) ar(BFE) = ar(AFD)
(e) ar(BFE) = 2*ar(FED)
(f) ar (FED) = 1/8 ar(AFC)

I got the diagram all right. I just need help solving it.

To the first answerer: E lies below BD, not CD.

Answer 1

Note that
1) Altitude of ABC is 2 times Altitude of BDE
2) BC is 2 times BD
3) BD is 3 times FD, or BF is 2 times FD

Using the triangle area formula A = 1/2 h b, where h is altitude and b is base length, it's not hard to derive (a) through (f), armed with facts 1), 2), and 3)

Answer 2

I'm not going to do all of these, but I'll help you set it up then give you an example.
If you have the diagram you're halfway there. Now, add some labels; (ABC) is equilateral so the sides are the same; label each on x. (BDE) is equilateral so the sides are the same, and you know one side is half the length of BC, so label these sides x/2.
Use Pythagorean theorem to find length of AD.
Notice that (ADC) and (CBE) are similar triangles and label the length of CE.
Notice that (ACE) is a right triangle, and use Pythagorean theorem to find the length of AE.
Notice that (ABED) is a trapezoid and use diagonals formula to label interior lengths.

Now just find and compare areas:
for (a):
ar(BDE) = 1/2 (x/2) (rt3 x/4) = (rt3 x^2) / 16
ar(ABC) = 1/2 (x) (rt3 x/2) = (rt3 x^2) / 4
As you can see, ar(ABC) is four times that of ar(BDE) so
ar(BDE) = 1/4 ar(ABC)

all the others can be done just the same

good luck

Answer 3

it is kind of hard for me to write all the details, but these points might help u:
since D is the midpoint, AD is a median and an altitude( a property of equilateral triangle), also AD is perpendicular to BC
Also area of equaliteral triangle can be expressed as ( u can check on this website) http://www.mathwords.com/a/area_equilateral_triangle.htm
or (side)^2* square root of 3/4
area of triangle is : (1/2)ABsinC, where A, B are sides of the trianlge and C is the angle enclosed between these sides.
Also, if we assume AB=a
then AD=(squareroot3)(a)/2
and BD=a/2
and angle ADE=150degree
i hope this information helps

Answer 4

I don't see how AE can intersect BD...unless somehow i drew my diagram wrong.

Answer 5

Erroneous question