Question

state and Prove the basic proportionality theorem using about solve the following in figure the heat is D is a point on the side BC of triangle ABC such that angle abc equal to angle BAC prove that c a b c d equal to c b by CA

Answer 1

Given: In triangleABC, DE is parallel to ABC.DG is perpendicular to BC. EF is perpendicular to AC.

To Prove:DE divides AC and BC in the same ratio.

Proof:Area (ADE) = (1/2) (AD) (EF).

Area (CDE) = (1/2) (DC) (EF)

Therefore Area (ADE) / Area (CDE) = AD / DC (1)

Area (BDE) = (1/2) (BE) (DG)

Area (CDE) = (1/2) (EC) (DG)

Area (BDE) / Area (CDE) = BE / EC (2)

But Area (BDE) = Area (ADE) since they are on the same base between the same parallels.

So (2) can be written as:Area (ADE) / Area (CDE) = BE / EC (3)From (1)

&

(3)AD / DC = BE / ECTherefore DE divides AC and BC in the same ratio.

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Answer 2

Step-by-step explanation:

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.