Question

in the adjoining figure,the side of AC of ∆ABC is produced to E such that CE = ½AC . If D is the midpoint of BC and ED produced meets AB at F, and CP,DQ are drawn parallel to BA, prove that FD = ⅓FE.​

Answer 1

$\huge\underline\bold\orange{Question}$

in the adjoining figure,the side of AC of ∆ABC is produced to E such that CE = ½AC . If D is the midpoint of BC and ED produced meets AB at F, and CP,DQ are drawn parallel to BA, prove that FD = ⅓FE.

$\huge\underline\bold\orange{Solution}$

In ∆ABC, D is the midpoint of BC and DQ || BA.

∴ Q is the midpoint of AC.

∴ AQ = QC.

Now, FA || DQ || PC, and AQC is there transversal such that AQ = QC and FDP is the other transversal on them

∴ FD = DP ⠀⠀⠀....(i) (by intercept theorem)

$Now \: EC = \frac{1}{2} AC = QC.$

∴In ∆EQD, C is the midpoint of EQ and CP || DQ.

∴ P must be the midpoint of DC

∴ DP = PE⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀.....(ii)

Thus, FD = DP = PE [from (i) and (ii)].

$Hence, \: FD = \frac{1}{3} FE$

━━━━━━━━━━━━━━━━━━━━━━━━━

$\sf{\underline{\underline{\blue{Additional\:Information-}}}}$

✯︎Intercept therom is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels.

✯︎the midpoint is the middle point of a line segment.

• It is equidistant from both endpoints, and it is the centroid both of the segment and of the endpoints.
• It bisects the segment.

━━━━━━━━━━━━━━━━━━━━━━━━━