It appears the angle equality is not even necessary $EF || CD$ if $\triangle{XCD}$ ~ $\triangle{XEF}$. This can be equivalently stated as \[\frac{XC}{XE} = \frac{XD}{XF}\] \[XC \times XF = XD \times XE\] Since $\triangle{XAF}$ ~ $\triangle{XCB}$, \[XC \times XF = XA \times XB\] Since $\triangle{XAD}$ ~ $\triangle{XEB}$ as well, \[XD \times XE = XA \times XB\] so the two are equal, as desired.

hyperbolictangent wrote: It appears the angle equality is not even necessary $EF || CD$ if $\triangle{XCD}$ ~ $\triangle{XEF}$. This can be equivalently stated as \[\frac{XC}{XE} = \frac{XD}{XF}\] \[XC \times XF = XD \times XE\] Since $\triangle{XAF}$ ~ $\triangle{XCB}$, \[XC \times XF = XA \times XB\] Since $\triangle{XAD}$ ~ $\triangle{XEB}$ as well, \[XD \times XE = XA \times XB\] so the two are equal, as desired. Please state what the point X is.

Assuming $X$ to be the point of intersection of the diagonals $AC$ and $BD$ of the quadrilateral, we have by BPT, $ \dfrac{XE}{XA} = \dfrac{XB}{XD} $ as $AD ||BE$ $ \dfrac{XA}{XC} = \dfrac{XF}{XB}$ as $BC||AF$. Multiplying the two equalities, we get $\dfrac{XE}{XC} = \dfrac{XF}{XD}$ By converse of BPT, $CD||EF$. Note that we neither require the angle equality nor the fact that the two angles are greater than $90^{\circ}$.

cmtappu96 wrote: Assuming $X$ to be the point of intersection of the diagonals $AC$ and $BD$ of the quadrilateral, we have by BPT, $ \dfrac{XE}{XA} = \dfrac{XB}{XD} $ as $AD ||BE$ $ \dfrac{XA}{XC} = \dfrac{XF}{XB}$ as $BC||AF$. Multiplying the two equalities, we get $\dfrac{XE}{XC} = \dfrac{XF}{XD}$ By converse of BPT, $CD||EF$. Note that we neither require the angle equality nor the fact that the two angles are greater than $90^{\circ}$. We don't require equality but I think $>90$ is used in the diagram to determine the direction of $E,F$

However, the direction of E,F does not matter since the BPT holds even if the points $E$ and $A$ are on the opposite sides of $X$ (and similarly even if $B$ and $D$ are opposite of $X$ and so on). I have attached a picture of the sketch I drew on Kig (I am too lazy to do the same again on the GeoGebra Applet )

Attachments:

Apply Pappus' Theorem on sets of collinear points $B-F-D$ and $A-E-C$ and since, $AF ||BC$, $AD ||BE$ $\implies$ $EF ||CD$

This problem can be solved solely using triangle similarity. I've posted my solution HERE in case anyone is interested, as well as many other geometry materials.