In the parallelogram $A B C D$, $B C=\lambda A B$. The internal angle bisectors starting from $A$ and $B$ intersect at point $M$. What fraction of the parallelogram $A B C D$ is covered by the triangle $A B M$?
Draw a line parallel to $AB$ through point $M$, and let this line intersect lines $BC$ and $AD$ at points $C^{*}$ and $D^{*}$, respectively. Let $\angle MAD^{*} = \angle MAB = \alpha$ and $\angle MBA = \angle MBC^{*} = \beta$. Notice that $\angle AMD^{*} = \angle MAB = \alpha$, since they are alternate interior angles, and similarly, $\angle BMC^{*} = \angle MBA = \beta$. Thus, triangles $\triangle AMD^{*}$ and $\triangle MBC^{*}$ are isosceles, from which it follows that $MD^{*} = D^{*}A = C^{*}B = MC^{*} = AB / 2$, where the second equality follows from the fact that $AB C^{*} D^{*}$ is a parallelogram, and the last equality from the fact that $MD^{*} + MC^{*} = D^{*}C^{*} = AB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_5d48156a890537c0bc32g-1.jpg?height=309&width=672&top_left_y=431&top_left_x=716)

Assume that $T_{AB C^{*} D^{*}} = 2$. Since $AB C^{*} D^{*}$ is a parallelogram and the side $AB$ and the corresponding height of $\triangle ABM$ are common, by the well-known area formulas, $T_{ABM} = 1$. Furthermore, by the given condition, $AD = \lambda AB = 2\lambda AD^{*}$, so again by the well-known area formula for a parallelogram,

\[
T_{ABCD} = AB \cdot AD \cdot \sin 2\alpha = 2\lambda (AB \cdot AD^{*} \cdot \sin 2\alpha) = 2\lambda T_{AB C^{*} D^{*}} = 4\lambda
\]

From the above, it also follows that point $M$ is inside the parallelogram $ABCD$ if and only if $\lambda \geq 1/2$. In this case, the desired area ratio is immediately given by:

\[
\frac{T_{ABM}}{T_{ABCD}} = \frac{1}{4\lambda}
\]

If $\lambda < 1/2$, then point $M$ lies outside $ABCD$. Let $AM$ and $BM$ intersect side $CD$ at points $D'$ and $C'$, respectively. In $\angle D^{*}AM$, by the intercept theorem for parallel lines,

\[
\frac{AD'}{AM} = \frac{AD}{AD^{*}} = 2\lambda
\]

It is clear that $\triangle MD'C' \sim \triangle MAB$, since their angles are equal in pairs, and the ratio of their similarity is $MD' / MA = 1 - DA' / MA = 1 - 2\lambda$. This implies that $T_{MD'C'} = (1 - 2\lambda)^2 T_{MAB} = (1 - 2\lambda)^2$. Finally, for the area covered by $\triangle ABM$,

\[
T_{AB C'D'} = T_{ABM} - T_{MC'D'} = 1 - (1 - 2\lambda)^2 = 4\lambda - 4\lambda^2
\]

From this, we get

\[
\frac{T_{AB C'D'}}{T_{ABCD}} = \frac{4\lambda - 4\lambda^2}{4\lambda} = 1 - \lambda
\]

Thus, the desired ratio is $1 / 4\lambda$ if $\lambda \geq 1/2$; and $1 - \lambda$ if $\lambda < 1/2$.