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Prove similarity theorem
CLASS 10
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Given : Two triangles ABC and DEF such thatAB BC CA
---- = ------ = ------
DE EF FD
Prove that : ΔABC ~ ΔDEF
Construction : Let P and Q be two points on DE and DF respectively such that DP = AB and DQ = = AC. Join PQ.
Statements Reasons
1) AB AC 1) Given
---- = ------
DE DF
2) DP DQ 2) As AB = DP and AC =DQ.
---- = ------ By substitution
DE DF
3) PQ || EF 3) By converse of basic
proportionality theorem
4) ∠DPQ = ∠E and ∠DQP = ∠F
4) Corresponding angles
5) ΔDPQ ~ ΔDEF 5) By AA similarity
6) DP PQ 6) By definition of similar triangles
---- = ------
DE EF
7) AB PQ
---- = ------ 7) As DP = AB , by substitution
DE EF
8) PQ BC 8) From (1) (6) and (7)
---- = ------
EF EF
9) PQ = BC 9) From (8)
10) ΔABC ≅ ΔDPQ 10) By S-S-S postulate
11) ΔABC ~ ΔDEF 11) From (5) and (10)
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anurag405093:
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