# Thread: Triangle. Equality of angles

1. ## Triangle. Equality of angles

Point M is the middle of the side BC of an acute-angles triangle ABC. Point K is on the side BC and $< BAM = < KAC$. On section AK there is a point E and $< BEK = < BAC$. Prove that $< KEC = < BAC$.

(Symbol "<" means an angle)

I would be very grateful for any help

2. Originally Posted by PaulinaAnna
Point M is the middle of the side BC of an acute-angles triangle ABC. Point K is on the side BC and $< BAM = < KAC$. On section AK there is a point E and $< BEK = < BAC$. Prove that $< KEC = < BAC$.

(Symbol "<" means an angle)
This is another tricky problem. The thing I found hard was to know how to make use of the fact that M is the midpoint of BC. Eventually, it occurred to me that this means that the triangles BAM, CAM have the same area (equal bases, same vertical height).

Denote by $\displaystyle \angle A,\ \angle B,\ \angle C$ the three angles of the triangle. Let $\displaystyle \alpha = \angle BAM = \angle KAC$ and let $\displaystyle \beta = \angle BAK = \angle MAC$.

From B, draw a line BX to meet AK at X, such that $\displaystyle \angle EBX = \angle B$. Then the triangles ABC, EBX are similar (because $\displaystyle \angle BEX = \angle A$ and $\displaystyle \angle EBX = \angle B$), and so $\displaystyle \dfrac{BE}{EX} = \dfrac{AB}{AC}$.

Also, $\displaystyle \angle BXA = \angle C$ and hence the points A, B, C, X are concyclic (having equal angles in the same segment). Therefore $\displaystyle \angle BXC = \angle B + \angle C$ (opposite angles of a cyclic quadrilateral are supplementary), and so $\displaystyle \angle EXC = \angle B$. Also, $\displaystyle \angle XBC = \angle KAC = \alpha$ and $\displaystyle \angle XCB = \angle BAK = \beta$ (angles in the same segment).

The area of triangle BAM is $\displaystyle \frac12AB.AM\sin\alpha$ and the area of triangle CAM is $\displaystyle \frac12AC.AM\sin\beta$. Since these are equal, it follows that $\displaystyle \dfrac{AB}{AC} = \dfrac{\sin\beta}{\sin\alpha}$.

Now look at the (vertical) distance from X to BC. This is equal to $\displaystyle BX\sin\alpha$ (since $\displaystyle \angle XBC = \alpha$). It is also equal to $\displaystyle CX\sin\beta$ (since $\displaystyle \angle XCB = \beta$). Therefore $\displaystyle \dfrac{BX}{CX} = \dfrac{\sin\beta}{\sin\alpha} = \dfrac{AB}{AC} = \dfrac{BE}{EX}$.

Putting together all the above facts, it follows by the RAR rule for similarity that the triangles EBX, EXC are similar. In particular, $\displaystyle \angle KEC = \angle BEX = \angle A$, as required.

3. Originally Posted by Opalg
Also, $\displaystyle \angle BXA = \angle C$
Where do we know this from?

4. Originally Posted by Glyper
Originally Posted by Opalg
Also, $\displaystyle \angle BXA = \angle C$
Where do we know this from?
This follows from the fact that the triangles ABC and EBX are similar. The angle at X in triangle EBX corresponds to the angle at C in triangle ABC.

5. Oh, thank you very much! I am stuck again, though:

Originally Posted by Opalg
the points A, B, C, X are concyclic (having equal angles in the same segment).
How do we know they're all concyclic and what equal angles in "the same segment" did you mean?