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 4 of 5: AN and KM are the perpendicular diagonals of AKNM

Previous step Next step

In plain words

$AKNM$ is really the same kite shape as $AKLM$ , just with its long diagonal stretched from $L$ out to $N$ — stretching a diagonal of a kite along its own line keeps the area formula $\tfrac12 d_1 d_2$ valid.

[AKNM] = \frac{1}{2}\cdot AN \cdot KM

Detailed analysis

Because $A$ , $L$ , $N$ lie on the same line (the bisector extended to the circle), the segment $AN$ lies along the same line as $AL$ , so $KM\perp AN$ as well as $KM\perp AL$ . Thus $AN$ and $KM$ are the two diagonals of quadrilateral $AKNM$ , and they are perpendicular, so its area is half their product: $[AKNM]=\tfrac12 AN\cdot KM$ .