Let $ABCD$ be a parallelogram ($AB < BC$). The bisector of the angle $BAD$ intersects the side $BC$ at the point K; and the bisector of the angle $ADC$ intersects the diagonal $AC$ at the point $F$. Suppose that $KD \perp BC$. Prove that $KF \perp BD$.
Let $BD\cap AK=T$
By angle bisector theorem $\frac{AF}{FC}=\frac{AD}{DC}=\frac{AD}{AB}=\frac{DT}{BT}$ $\implies$ $TF||BK||AD$ now let $\angle DBK=\beta=\angle FTD$ and $\angle BAK=\angle AKB=\angle FDK=\alpha$ (simple angle chasing).Using sine law in triangles $TFD$ and $TKD$ get $\frac{FD}{TD}=\frac{sin\beta}{cos\alpha}$ and $\frac{TK}{TD}=\frac{cos\beta}{cos\alpha}$ $\implies$ $\frac{FD}{TK}=tg\beta=\frac{DK}{BK}$ and also using $\angle FDK=\angle TKB=\alpha$ get $\Delta BKT\sim\Delta KDF$ $\implies$ $\angle FKD=\angle KBD=\beta$ $\implies$ $KF\perp BD$
