In the figure ABCD is a trapezium with AB || DC. E and F are points on non-parallel sides AD and BC respectively such that EF is parallel to AB. Prove that $\frac{AE}{ED} = \frac{BF}{FC}$.
In the above figure, if $AD = 2.0$ cm, $AE = 1.8$ cm, $EC = 3.6$ cm and $DE \parallel BC$ the measure of $BD$ will be :
In the figure, in $\triangle MNL$ and $\triangle PQR$, $\angle M = \angle Q = 70^\circ$, $MN = 3$ cm, $ML = 4.5$ cm, $PQ = 2$ cm, and $QR = 3$ cm. Then, the following correct relation will be:
In the figure, if \(DE \parallel BC\), then the measure of \(CE\) will be:
If \(\dfrac{AO}{OC} = \dfrac{BO}{OD} = \dfrac{1}{2}\) and \(AB = 5 \, \text{cm}\) in the following figure, then find the value of \(DC\).
In the figure, there is a pair of similar triangles. Its correct symbolic expression will be
In the given figure, if \( ST \parallel QR \), \( QS = 3 \text{ cm} \), \( SR = 1.5 \text{ cm} \), and \( PT = 2.8 \text{ cm} \), then find the value of \( TR \).
In the figure, if $LM \parallel CB$ and $LN \parallel CD$, then prove that $\dfrac{AM}{BM} = \dfrac{AN}{DN}$.