PROOF OF BPTGiven: In ΔABC, DE is parallel to BCLine 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=AECEConstruction: Join segments DC and BEProof: 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=AECEHence Proved. This is base proportionality theorem. ??
If in triangli one line intersect each sides theh this triangle are divide in ratio whic are proptional to eaxh other