Answer the Q.28
please answer it fast ...AP is not constructed it is done by mistake
Attachments:
Answers
Answered by
1
Answer:
Step-by-step explanation:
Attachments:
Answered by
0
Given: ABCD is a parallelogram, E and F are mid points of AB and DC and a line GH intersects AD, EF and BC at G, P, and H respectively.
Construction: Join GB and Let it intersect EF of X.
Now since E and F are mid points of AB and DC
⇒ AE = EB = AB
and DF= FC = DC
but AB = DC
⇒ AEFD and BCFE are parallelogram
⇒ AD EF BC ........ (1)
Now In ∆ ABG
AE = EB and EX AG (from (1))
⇒ X is the mid point of GB (mid point theorem)
and in ∆ GBH
GX = XB (∵ X is the mid point of GB)
and XP BH (from (1))
⇒ P is the mid point of GH
⇒ GP=PH
Thus proved !!!
Similar questions