Suppose that $ABCD$ is a trapezoid with $AB$ parallel to $CD$. Let $P$ be the point where the diagonals $AC$ and $BD$ intersect. Show that the triangles $CDP$ and $ABP$ are similar. Use this to prove that the diagonals of a trapezoid cut each other into segments that are proportional to the parallel sides of the trapezoid.
I am having trouble showing that $CDP$ and $ABP$ are similar. Also, i am not sure where to start with showing the diagonals of a trapezoid cut each other into segments that are proportional to the parallel sides of the trapezoid. Here is what I have gotten so far
$\textbf{Attempt:}$
If two triangle are similar than it follows that one triangle must be a scaler of the other triangle. This can be shown by establishing linear dependence. Recall from linear algebra that a system is linearly dependent when the determinant is equal to zero. Therefore, see that \begin{align*} \det\begin{pmatrix} a & d & 1\\ b & c & 1\\ p&p&1 \end{pmatrix}=0&\Rightarrow\det\begin{pmatrix} a & d&1\\ b-a&c-d&0\\ p-a&p-d&0 \end{pmatrix}=0\\ &\Rightarrow(b-a)(p-d)-(p-d)(c-d)=0\\ &\Rightarrow(b-a)(p-d)=(p-d)(c-d)\\ &\Rightarrow\dfrac{|p-a|}{|b-a|}=\dfrac{|p-d|}{|c-d|}\\ &\Rightarrow\dfrac{|PA|}{|BA|}=\dfrac{|PD|}{|CD|} \end{align*} We must now show this holds for our given triangles. . . I have tried using sine law for this but it doesn't seem to work. Any nudges in the right direction are much appreciated.
Continuing from my hint, here's the proof. (Don't peek, Jeremy, until you've read my hint.)
$\angle APB=\angle CPD$, because they're a pair of vertical angles (opposing angles formed when two lines intersect). https://en.wikipedia.org/wiki/Angle#Vertical_and_adjacent_angle_pairs
Also, if $L_1$ and $L_2$ are parallel lines cut by a transversal $L_3$ (a line not parallel to $L_1$), then the alternate interior angles are equal. https://en.wikipedia.org/wiki/Transversal_(geometry)#Alternate_angles
Applying this fact to the lines $AB$ and $CD$, as cut by the transversal $BD$, $\angle ABP=\angle CDP$. Applying this fact to the same parallel lines cut by $AC$, $\angle PAB = \angle PCD$.
Hence $(\angle APB,\angle ABP, \angle PAB)=(\angle CPD, \angle CDP, \angle PCD)$, making the triangles $APB$ and $CPD$ similar. QED. https://en.wikipedia.org/wiki/Similarity_(geometry)#Similar_triangles