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Bpt theorem from chapter 6

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Bpt theorem from chapter 6

Posted by Kunal Kunal 6 years, 10 months ago

CBSE > Class 10 > Mathematics

2 answers

Avinash Saigal 6 years, 10 months ago

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. This is base proportionality theorem. ??

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Kratika Jhala 6 years, 10 months ago

If in triangli one line intersect each sides theh this triangle are divide in ratio whic are proptional to eaxh other

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Posted by Hari Anand 6 months, 2 weeks ago

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Don't post personal information, mobile numbers and other details.
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Posted by Kanika . 1 month, 1 week ago

CBSE > Class 10 > Mathematics