Question

state and prove bpt theorem when construction is DN perpendicular to AE and EM perpendicular to AD​

Answer 1

Answer:

BPT theorem states that if a line is drawn parallel to one side of a triangle to intersect the other 2 sides in distinct points, the other two sides are divided in the same ratio.

Given:

A Triangle ABC with DE parallel to BC

To Prove:

$\frac{AD}{BD} = \frac{AE}{EC} </p><p>$

Construction:

Construct DN perpendicular to AE and EM perpendicular to AD.Join BE and DC

Proof:

Consider ADE , AE is the base and DN is the height

Therefore,

$Ar (ADE)=1/2×AE×DN (1)$

Similarly, considering AD as the base and EM as the height

$Ar (ADE)=1/2×AD×EM (2)$

Now, Consider BDE

$Ar (BDE)=1/2×BD×EM(3)$

Now, Consider CDE

$Ar (CDE)=1/2×CE×DN(4)$

Dividing (1) by (4) we get,

$\frac{Ar (ADE)=1/2×AE×DN}{Ar (CDE)=1/2×CE×DN} = \frac{AE}{AE} \: \: \: \: {5}$Similarly (2)÷(3) we get,

$\frac{Ar (ADE)=1/2×AD×EM }{Ar (BDE)=1/2×BD×EM } = \frac{AD}{BD } \: \: (6)$

Now,BDE and DEC are on the same base DE and between the same parallels BC and DE

So,

$ar(BDE)= ar(DEC) \: \: (7)$

From (5), (6), (7) we can conclude that

$\frac{AD}{BD} = \frac{AE}{EC}$

HENCE PROVED

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Answer 2

Answer:-

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: => AD/DB = AE/AC

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)

= ½ ×AD×EF / ½ ×DB×EF

= AD/DB ......(1)

In ΔADE and ΔCDE,

=> Ar(ADE)/Ar(ECD)

= ½×AE×DG / ½×EC×DG

= AE/EC ........(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,

=> AD/DB = AE/AC

Hence Proved.

i hope it helps you.