In the adjoining figure, ABCD is a parallelogram. E is mid-point of DC and F is a point on AC such that CF = (1/4)AC. If EF produced meets BC at G, prove that:
(i) G is mid-point of BC
(ii) 2EG = DB.
Answers
Answer:
Hey buddy,here is your answer from the deepest point of hell.....XD
Given: ABCD is a parallelogram. E is the mid point of CD.
Q is a point on AC such that CQ=(1/14)AC
EQ produced meet BC in R.
To prove : R is the mid point of BC
Construction : join BD in O.Let BD intersect AC in O.
Prove : O is the mid point of AC. {diagnols of parallelogram bisect each other }
∴ OC = (1/2) AC
=> OQ = OC-CQ = (1/2)AC - (1/4)AC = (1/4)AC.
=> OQ = CQ
∴ Q is the mid point of OC.
In triangle OCD,
E is the mid point of CD and Q is the mid point of OC,
therefore EQ is parallel to OD (Mid point theorem)
=> ER is parallel to BD
In traingle BCD,
E is the midpoint of CD and ER is parallel to BD,
∴ R is the mid point of BC (Converse of mid point theorem) (Proved)
Explanation: