Try this beautiful problem from Geometry based on Area of a Triangle Using similarity
In $\triangle ABC $ , a point E is on AB with AE = 1 and EB=2.Point D is on AC so that DE $\parallel$ BC and point F is on BC so that EF $\parallel$ AC.
What is the ratio of the area of quad. CDEF to the area of $\triangle ABC$?
Geometry
Area
similarity
Answer:$\frac{4}{9}$
AMC-8, 2018 problem 20
Pre College Mathematics
$\triangle ADE$ $\sim$ $\triangle ABC$
Can you now finish the problem ..........
$\triangle BEF$ $\sim$ $\triangle ABC$
can you finish the problem........
Since $\triangle ADE$$\sim$ $\triangle ABC$
$\frac{ \text {area of} \triangle ADE}{ \text {area of} \triangle ABC}$=$\frac{AE^2}{AB^2}$
i.e $\frac{\text{area of} \triangle ADE}{\text{area of} \triangle ABC}$ =$\frac{(1)^2}{(3)^2}$=$\frac{1}{9}$
Again $\triangle BEF$ $\sim$ $\triangle ABC$
Therefore $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$=$\frac{BE^2}{AB^2}$
i.e $\frac{ \text {area of} \triangle BEF}{ \text {area of} \triangle ABC}$ =$\frac{(2)^2}{(3)^2}$=$\frac{4}{9}$
Therefore Area of quad. CDEF =$\frac {4}{9}$ of area $\triangle ABC$
i.e The ratio of the area of quad.CDEF to the area of $\triangle ABC$ is $\frac{4}{9}$
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