Question

bpt theorem proving​

Answer 1

Answer:

Basic Proportionality theorem was introduced by a famous Greek Mathematician, Thales, hence it is also called Thales Theorem. According to him, for any two equiangular triangles, the ratio of any two corresponding sides is always the same. Based on this concept, he gave theorem of basic proportionality. This concept has been introduced in similar triangles.  If two triangles are similar to each other then,

i) Corresponding angles of both the triangles are equal

ii) Corresponding sides of both the triangles are in proportion to each other

Thus two triangles ΔABC and ΔPQR are similar if,

i) ∠A=∠P, ∠B=∠Q and ∠C=∠R

ii) AB/PQ, BC/QR, AC/PR

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Step-by-step explanation:

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Answer 2

BPT

If a line is parallel to a side of a triangle which intersects the other sides into two distinct points, then the line divides those sides in proportion. 

Given: In  ΔABC, DE is parallel to BC

Given: In  ΔABC, DE is parallel to BCLine DE intersects sides AB and AC in points D and E respectively.

Given: In  ΔABC, DE is parallel to BCLine DE intersects sides AB and AC in points D and E respectively.To Prove: ADBD=AECE

Given: In  ΔABC, DE is parallel to BCLine DE intersects sides AB and AC in points D and E respectively.To Prove: ADBD=AECEConstruction: Draw EF ⟂ AD and DG⟂ AE and join the segments BE and CD.

Proof 

 Area of Triangle= ½ × base × height

 Area of Triangle= ½ × base × heightIn ΔADE and ΔBDE,

 Area of Triangle= ½ × base × heightIn ΔADE and ΔBDE,Ar(ADE)Ar(DBE)=12×AD×EF12×DB×EF=ADDB(1)

 Area of Triangle= ½ × base × heightIn ΔADE and ΔBDE,Ar(ADE)Ar(DBE)=12×AD×EF12×DB×EF=ADDB(1)In ΔADE and ΔCDE,

 Area of Triangle= ½ × base × heightIn 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=AE