verification of basic proportionality theorem.
Answers
Answered by
12
PROOF OF BPT
Given: In ΔABC, DE is parallel to BC
Line DE intersects sides AB and PQ in points D and E, such that we get triangles A-D-E and A-E-C.
To Prove: ADBD=AECE
Construction: Join segments DC and BE
Proof:
In ΔADE and ΔBDE,
A(ΔADE)A(ΔBDE)=ADBD (triangles with equal heights)
In ΔADE and ΔCDE,
A(ΔADE)A(ΔCDE)=AECE (triangles with equal heights)
Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,
A(ΔBDE)=A(ΔCDE)
Therefore,
A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)
Therefore,
ADBD=AECE
Hence Proved.
Given: In ΔABC, DE is parallel to BC
Line DE intersects sides AB and PQ in points D and E, such that we get triangles A-D-E and A-E-C.
To Prove: ADBD=AECE
Construction: Join segments DC and BE
Proof:
In ΔADE and ΔBDE,
A(ΔADE)A(ΔBDE)=ADBD (triangles with equal heights)
In ΔADE and ΔCDE,
A(ΔADE)A(ΔCDE)=AECE (triangles with equal heights)
Since ΔBDE and ΔCDE have a common base DE and have the same height we can say that,
A(ΔBDE)=A(ΔCDE)
Therefore,
A(ΔADE)A(ΔBDE)=A(ΔADE)A(ΔCDE)
Therefore,
ADBD=AECE
Hence Proved.
Answered by
5
☺❤☺❤☺❤☺HEY MATE YOUR ANSWER IS HERE. ..☺❤☺❤☺❤☺
Basic Proportionality Theorem was first stated by Thales, a Greek mathematician. ... Basic Proportionality Theorem (can be abbreviated as BPT) states that, 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
⭐⭐⭐⭐HOPE IT'S HELPFUL. .⭐⭐⭐⭐⭐⭐
☺☺☺☺☺PLZZZ MARK ME BRAINLIEST ☺☺☺☺☺☺
Basic Proportionality Theorem was first stated by Thales, a Greek mathematician. ... Basic Proportionality Theorem (can be abbreviated as BPT) states that, 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
⭐⭐⭐⭐HOPE IT'S HELPFUL. .⭐⭐⭐⭐⭐⭐
☺☺☺☺☺PLZZZ MARK ME BRAINLIEST ☺☺☺☺☺☺
Attachments:
bebita:
thnk u but i want the verify not defination....☺☺☺
Similar questions