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 3 of 5: Compute the diagonal KM of the kite

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In plain words

The kite's short diagonal $KM$ is twice the altitude of the right triangle $AKL$ onto its hypotenuse $AL$ — a quantity that collapses neatly to $AL\sin A$ via the double-angle formula.

KM = 2\cdot\frac{AK\cdot KL}{AL} = 2\cdot AL\cos\frac{A}{2}\sin\frac{A}{2} = AL\sin A

Detailed analysis

Let $Q$ be the intersection of the diagonals $AL$ and $KM$ . Since $AKLM$ is a kite, $Q$ is the foot of the altitude from $K$ (and from $M$ ) to $AL$ in the right triangle $AKL$ , and $KQ$ (half of $KM$ ) equals $(AK\cdot KL)/AL$ by the standard altitude-on-hypotenuse relation. Substituting $AK=AL\cos(A/2)$ , $KL=AL\sin(A/2)$ from Step 1 gives $KM = 2AL\sin(A/2)\cos(A/2) = AL\sin A$ .