explain about similar triangles....
bpt...
converse of bpt...
how to find....
Answers
similar triaingles be find by -
(1)side side side
(2)side angle side
(3)angle angle side
bpt - parallel lines
PROOF OF BPT
Given: In ΔABC, DE is parallel to BC
Line DE intersects sides AB and AC in points D and E respectively.
To Prove: ADBD=AECE
Construction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.
Proof:
Area of Triangle= ½ × base× height
In ΔADE and ΔBDE,
Ar(ADE)Ar(DBE)=12×AD×EF12×DB×EF=ADDB(1)
In ΔADE and ΔCDE,
Ar(ADE)Ar(ECD)=12×AE×DG12×EC×DG=AEEC(2)
Note that ΔDBE and ΔECD have a common base DE and lie between the same parallels DE and BC. Also, we know that triangles having the same base and lying between the same parallels are equal in area.
So, we can say that
Ar(ΔDBE)=Ar(ΔECD)
Therefore,
A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)
Therefore,
ADBD=AECE
Hence Proved.
The BPT also has a converse which states, if a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
