Show that $AB = 3AD$

Question

Given that $AF=EF$ and $BE=CE$. Show that $AB=3AD$.

This question was given during my exams today and it surprised my whole class. No one knew how to start and any tips would be helpful!

Answer 1

Let $G$ be a midpoint of $BD$.

Thus, $GE||DC$ and since $F$ is a midpoint of $AE$, we obtain $AF=FG=GB$ and we are done!

Answer 2

Use Menelaus's Theorem for $\Delta ABE$ and line $CFD$:

$$\frac{CB}{CE}\cdot\frac{FE}{FA}\cdot\frac{DA}{DB}=1$$

but $AF=EF$ and $BE=CE$, so,

$$2\cdot1\cdot\frac{DA}{DB}=1\to DB=2DA \to AB=3AD$$

Answer 3

Let $X$ be a point on $BD$ such that $EX\parallel CD$. Then $$\frac{BX}{XD}=\frac{BE}{EC}=1$$ and $$\frac{AD}{DX}=\frac{AF}{FE}=1.$$ It follows that $AD=DX=XB$, so $AB=3AD$.

Answer 4

Draw $F'$ symmetric of $F$ wrt $E$

$BFCF'$ is a parallelogram and for Thales' theorem as $FF'=2AF$ also $BD=2AD$

QED