# Question #a40d8

Dec 28, 2016

see explanation.

#### Explanation:

Let $\angle E A D = x , \angle A D C = y$
Given $A E = E F , \implies \angle E F A = x , \implies \angle B F D = x$

Sine Law :

$\Delta A C D , \frac{C D}{\sin} x = \frac{A C}{\sin} y - - - - - \left(1\right)$
$\Delta B F D , \frac{D B}{\sin} x = \frac{B F}{\sin} \left(180 - y\right) = \frac{B F}{\sin} y - - - - - \left(2\right)$

Given $C D = D B , \frac{C D}{\sin} x = \frac{D B}{\sin} x , \implies \left(1\right) = \left(2\right)$
$\implies \frac{A C}{\sin} y = \frac{B F}{\sin} y , \left(\sin \left(180 - y\right) = \sin y\right)$

Hence, $A C = B F$ (proved)

Dec 28, 2016

Given ABC a triangle where [AD] is the median and let the segment line [BE] which meets [AD] at F and [AC] at E If we assume that AE=EF, we are to show that AC=BF.

Construction

AD is extended up to G such that $A D = D G$ and the point G is joined with C,D and B.

Proof

Now in quadrilateral $A B G C$, D the point of intersection of two diagonals is the mid point of the diagonals as $C D = D B \left(\text{given");AD =DG("by construction}\right)$ So quadrilateral $A B G C$ is a parallelogram.

In $\Delta A E F$
$A F = F E \to \angle F A E = \angle A F E = \angle B F G \left(\text{vertically opposite}\right)$
$\implies \angle C A G = \angle B F G$

Again $A C | \setminus | B G$ and AG intercept
So $\angle C A G = \text{alternating} \angle A G B = \angle B G F$

So $\angle B F G = \angle B G F \to B F = B G$
Again $B G = A C \left(\text{opposite sides of a parallelogram}\right)$

Hence $A C = B F$