In a triangle, ABC, D is a point on AB such that AB = 4AD and E is a point on AC such that AC = 4AE. Prove that BC = 4ED.
Answer:
- It is given in the question that D and E are the points on side AB and BC of ΔABC respectively. Join DE.
- Given:- AB = 4AD
or, AD =
AB1 4
AC = 4AE
or, AE =
AC1 4 - We need to prove that the ΔADC and ΔABC are similar.
Where, A is the common angle in ΔADE and ΔABC.
Therefore,
=AD AB
[By BPT theorem.]AE AC
Then,
=AD AB
=AE AC
------(1)ED BC
So, ΔABC ∼ ΔADE [By SAS criteria.] - As,
=AD AB
------(2)1 4
and
=AE AC
------(3)1 4 - On comparing (1), (2) and (3), we get:
=ED BC 1 4 - Hence, BC = 4ED.