Define vertically opposite angles.

Prove to corresponding sides of similar triangles

Class 9, EXERCISE 8.2, Page No: 151, Questions No:5.

Session 2023-2024 are based on CBSE Board.

Share

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

In quadrilateral ABCD,

AB = CD [∵ Given]

(1/2)AB = (1/2)CD

⇒ AE = CF [∵ E and F are the mid-points of AB and CD respectively]

In Quadrilateral AECF,

AE = CF [∵ Proved above]

AE ∥ CF [∵ Opposite sides of a parallelogram]

Hence, AECF is a parallelogram.

In ΔDCQ,

F is mid-point of DC [∵ Given]

and FP ∥ CQ [∵ AECF is a Parallelogram]

Hence, P is mid-point of DQ [ Converse of mid-point theorem]

Hence, DP = PQ …(1)

Similarly,

In ΔABP,

E is mid-point of AB [∵ Given]

and EQ ∥ AP [∵ AECF is a parallelogram]

Hence, Q is mid-point of PB [∵ Converse of mid-Point Theorem]

Hence, PQ = QB …(2)

From (1) and (2), we have

DP = PQ = QB

Hence, line segment AF and EC trisect BD.