If ADBD=35, AC = 14 cm and DE∥BC Find EC.
8.75
It is given that ADBD = 35
∵DE∥BC,In △ADE and △ABC,
∠ADE = ∠ABC [corresponding angles]
∠DAE=∠BAC [common angle]
Since two angles are equal, it follows that the third angle is also correspondingly equal and hence the triangles are similar.
∴ADBD=AEEC=35
ECAC=58
EC=58×AC
EC=58×14=8.75cm