The side AC of a triangle ABC is produced to E such that CE = 1/2 AC. If D is midpoint of BC and ED produced meets AB in F, and CP, DQ are drawn parallel to BA. Prove that FD = 1/3 EF

Asked by Paresh | 8th Dec, 2015, 04:26: PM

Expert Answer:

In space triangle ABC DQ space vertical line vertical line vertical line space space BA rightwards double arrow DO space vertical line vertical line vertical line space space BA By space using space BPT space theorem space we space can space say space that CD over DB equals CO over OA But space straight D space is space midpoint space of space BC So space DB space equals space CD  DB over DB equals CO over OA 1 equals CO over OA OA space equals CO  AC space equals OA space plus OC space equals space OA plus OA rightwards double arrow OA equals 1 half AC rightwards double arrow OA equals CO equals 1 half AC equals CE space space space space space space left square bracket Given space CE space equals space 1 half AC space right square bracket  In space triangle AEF DQ space vertical line vertical line vertical line space space BA rightwards double arrow DO space vertical line vertical line vertical line space space FA  By space using space BPT space theorem space we space can space say space that EO over OA equals ED over DF  fraction numerator OC plus CE over denominator OA end fraction equals ED over DF  fraction numerator 2 OA over denominator OA end fraction equals ED over DF 2 DF space equals space ED  EF space equals space FD plus ED EF space equals space FD plus 2 FD EF space equals space 3 FD FD space equals 1 third EF  space
 

Answered by Vijaykumar Wani | 9th Dec, 2015, 11:32: AM