Question

D, E, F are the mid-points of the sides BC, CA and AB of a triangle ABC. FG is drawn parallel to BE, meeting DE produced in G. Prove that the sides

of the triangle CFG are equal to the medians of the triangle ABC.
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Answer 1

QuesTion ⤵

D, E, F are the mid-points of the sides BC, CA and AB of a triangle ABC. FG is drawn parallel to BE, meeting DE produced in G. Prove that the sides of the triangle CFG are equal to the medians of the triangle ABC.

AnsWer ⬇

Solution refer to the attachment .

Formula used : BPT or Thales theorem .

Thanks...

Answer 2

Answer:

Solution:

In the given question, we know Triangle DEF formed with midpoints is similar to the Outer Triangle ABC

On the basis of the similarity, we can say,

If two triangles are similar then the ratio of their area is equal to the square of the ratio of their corresponding sides

Mathematically can be written as :-

$\frac{\text {area of triangle } D E F}{\text {area of triangle ABC}}=\left(\frac{D E}{A C}\right)^{2} \:$

Since, DECF is a parallelogram. So DE = FC

On substituting:

$\frac{\text {area of triangle } D E F}{\text {area of triangle ABC}}=\left(\frac{F C}{A C}\right)^{2} \:$

Also, F is the midpoint of AC

$\text { So } \mathrm{AC}=2 \times \mathrm{FC} \:$

$\begin{lgathered}\begin{array}{l}{\frac{\text {area of triangle } D E F}{\text {area of triangle } A B C}=\left(\frac{F C}{2 F C}\right)^{2}} \\\\ {\frac{\text {area of triangle } D E F}{\text {area of triangle } A B C}=\frac{1}{4}}\end{array}\end{lgathered} \:$

Hence ratio of area of triangle DEF and triangle ABC is given as:

Ratio of area of triangle DEF :  area of triangle ABC = 1 : 4

hope this helps you