Question

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Answer 1

Answer:

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

Answer 2

Given :- DE // BC, AD = 1cm and BD = 2cm

Now, in triangle ADE

AD = 1cm

And, in triangle ABC

AB = AD + BD

AB = 1 + 3

AB = 4

Now, Ratio of area of triangle ABC to the area of triangle ADE is :

ar( ABC )/ ar ( ADE ) = ( AB ) ² / ( AD ) ²

= 4²/1²

= 16/1

Hence the ratio is 16:1