Problem 2

In an acute-angled triangle

ABC

, the interior bisector of angle

A

intersects

BC

L

and intersects the circumcircle of

ABC

again at

N

. From point

L

, perpendiculars are drawn to

AB

and

AC

, with feet

K

and

M

respectively. Prove that the quadrilateral

AKNM

and the triangle

ABC

have equal areas.

Step 1 of 5: AKLM is a kite: perpendicular diagonals

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In plain words

Folding the figure along $AL$ swaps $K$ with $M$ and $B$ 's side with $C$ 's side, because $AL$ is exactly the angle bisector — this symmetry is what forces $KM\perp AL$ .

AK = AM = AL\cos\frac{A}{2}, \qquad LK = LM = AL\sin\frac{A}{2}

Detailed analysis

Since $AL$ bisects angle $A$ and $LK \perp AB$ , $LM \perp AC$ , the right triangles $AKL$ and $AML$ share the hypotenuse $AL$ and the equal angle $\angle KAL = \angle MAL = A/2$ , so they are congruent: $AK=AM=AL\cos(A/2)$ and $LK=LM=AL\sin(A/2)$ . Hence $AKLM$ is a kite symmetric about $AL$ , and its diagonals $AL$ and $KM$ are perpendicular.