BD and CE intersect each other at the point P
Is ΔPBC similar to ΔPDE ? Why ?
D
C
B
E
5 cm
12 cm
10 cm
6 cm
P
Proof :
BP
DP
=
5
10
BP
DP
=
1
2
∴
…(i)
CP
PE
=
6
12
CP
PE
=
1
2
∴
…(ii)
In ΔPBC and ΔPDE
BP
DP
=
CP
PE
[From (i) and (ii)]
∠BPC
=
∠DPE
[Vertically opposite angles]
ΔPBC
~
ΔPDE
(SAS criterion)
P
∴
Example
K
L
M
P
N
a
x
46o
46o
b
c
Express x in term of a, b and c.
Sol:
In ΔPKN and ΔLKM
∠K
=
∠K
(Common angle)
∠KNP
=
∠KML
(Given)
ΔPKN
~
ΔLKM
∴
(AA criterion)
KN
KM
PN
LM
=
(Corresponding sides of similar triangles)
b
(b + c)
x
a
=
x
=
ab
b + c
(
)
Example