ABCD is a parallelogram. E and F are mid points of BC and
CD respectively. AE and AF meet diagonal BD in Q and P
respectively. Show that P and Q trisect BD.
Answers
Step-by-step explanation:
Let A, B, C, D, E, F, PQ have position vectors
a
=
,
b
ˉ
,
c
ˉ
,
d
ˉ
,
e
ˉ
,
f
ˉ
,
p
ˉ
,
q
ˉ
respectively.
becauseABCD is a parallelogram
∴
AB
=
DC
∴
b
ˉ
−
a
ˉ
=
c
ˉ
−
d
ˉ
∴
c
ˉ
=
b
ˉ
+
d
ˉ
−
a
ˉ
.....(1)
E is the midpoint of BC
∴
e
ˉ
=
2
b
ˉ
+
c
ˉ
∴E is the midpoint of BC
∴
e
ˉ
=
2
b
ˉ
+
c
ˉ
∴2
e
ˉ
=
b
ˉ
+
c
ˉ
.....(2)
f is the midpoint of CD
∴
f
ˉ
=
2
c
ˉ
+
d
ˉ
∴2
f
ˉ
=
c
ˉ
+
d
ˉ
.....(3)
2
e
ˉ
=
b
ˉ
+
c
ˉ
.....By (2)
=
b
ˉ
+(
b
ˉ
+
d
ˉ
+
a
ˉ
.....by (1)
∴2
e
ˉ
+
a
ˉ
=2
b
ˉ
+
d
ˉ
∴
2+1
2
e
ˉ
+
a
ˉ
=
2+1
2
b
ˉ
+
d
ˉ
LHS is the position vector of the point on AE and RHS is the position vector of the point on DB. But AE and DB meet at Q.
∴
q
ˉ
=
2+1
2
b
ˉ
+
d
ˉ
∴Q divides DB in the ratio 2:1 ......(4)
2
f
ˉ
=
c
ˉ
+
d
ˉ
........[By (3)]
=(
b
ˉ
+
d
ˉ
−
a
ˉ
)+
d
ˉ
.....(1)
∴2
f
ˉ
+
a
ˉ
=2
d
ˉ
+
b
ˉ
∴
1+2
a
ˉ
+2
f
ˉ
=
1+2
b
ˉ
+2
d
ˉ
LHS is the position vector of the point on AF and RHS is the position vector of the point on DB. But AF and DB meet at P.
∴
p
ˉ
=
1+2
b
ˉ
+2
d
ˉ
∴P divides DB in the ratio 1:2 .........(5)
From (4) and (5), if follows that P and Q trisect DB.