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●Prove Converse Of BPT THEOREM

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Answer 1

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Converse of Basic Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

If AD AE

---- = ------ then DE || BC

DB EC

Given : A Δ ABC and a line intersecting AB in D and AC in E,

such that AD / DB = AE / EC.

Prove that : DE || BC

Let DE is not parallel to BC. Then there must be another line that is parallel to BC.

Let DF || BC.

Statements

Reasons

1) DF || BC 1) By assumption

2) AD / DB = AF / FC 2) By Basic Proportionality theorem

3) AD / DB = AE /EC 3) Given

4) AF / FC = AE / EC 4) By transitivity (from 2 and 3)

5) (AF/FC) + 1 = (AE/EC) + 1 5) Adding 1 to both side

6) (AF + FC )/FC = (AE + EC)/EC 6) By simplifying

7) AC /FC = AC / EC 7) AC = AF + FC and AC = AE + EC

8) FC = EC 8) As the numerator are same so denominators are equal

This is possible when F and E are same. So DF is the line DE itself.

∴ DF || BC

Examples

1) D and E are respectively the points on the sides AB and AC of a ΔABC such that AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm, prove that DE || BC.

Solution :

AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm

∴ BD = AB – AD = 5.6 – 1.4 = 4.2 cm

And EC = AC – AE = 7.2 – 1.8 = 5.4 cm

Now, AD / DB =1.4 / 4.2 = 1/3

And AE / EC = 1.8 / 5.4 = 1/3

⇒ AD / DB = AE / EC

DE || BC ( by converse of basic proportionality theorem)

---------------------------------------------------------------

2) State whether PQ || EF.

DP / PE = 3.9 / 3 = 13/10

DQ / QF = 3.6 /2.4 = 3/2

So, DP / PE ≠ DQ / QF

⇒ PQ ∦ EF ( PQ is not parallel to EF)

Answer 2

Step-by-step explanation:

Converse of BPT(Basic Proportionality Theorem).

In ΔABC, DE is a straight line such that AD/DB = AE/EC.

Prove: DE ║ BC.

Proof:

Consider a line DE such that (AD/DB) = (AE/EC).

Draw a line DE' parallel to BC(If DE is not parallel to BC).

⇒ (AD/DB) = AE'/E'C     ----------- (1)

In ΔABC,

⇒ (AD/DB) = AE/EC     ---------- (2)

From (1) & (2), we get

⇒ (AE/EC) = AE'/E'C

⇒ (AE/EC) + 1 = (AE'/E'C) + 1

⇒ (AE + EC)/EC = (AE' + E'C)/E'C

⇒ AE/EC = AC/E'C

⇒ EC = E'C

∴ DE ║ BC.

Hope it helps!