In $\triangle \mathrm{ABC}, \mathrm{D}$ is the mid-point of side $\mathrm{AC}$ and $\mathrm{E}$ is a point on side $\mathrm{AB}$ such that $\mathrm{EC}$ bisects $\mathrm{BD}$ at $\mathrm{F}$. If $\mathrm{AE}=30 \mathrm{~cm}$, then the length of EB is:
Answer & explanation
Correct answer: option 4
Draw line parallel to EC from point D on side AE at G.
DG is parallel to EC
In \(\Delta \)AEC, DG is parallel to EC
= \(\frac{AG}{GE}\) = \(\frac{AD}{DC}\)
= AG = GE as AD = DC
Similarly, GD is parallel to EF for \(\Delta \)BGD,
= \(\frac{EB}{EG}\) = \(\frac{BF}{FD}\)
= EB = GE as BF = BD
= AE = AG + GE = 30
= AG = GE = 15
Therefore, EB is 15 cm.