Prove the AAA criteria Class 10 Triangles
Answers
Answer:
Given : Triangles ABC and DEF such that ∠A = ∠D; ∠B = ∠E; ∠C = ∠F
Prove that : Δ ABC ~ ΔDEF
Construction : We mark point P on the line DE and Q on the line DF such that AB = DP and AC = DQ, we join PQ.
There are three cases :
Case ( i ) : AB = DE, thus P coincides with E.
Statements
Reasons
1) AB = DE 1) According to 1st case
2) ∠A = ∠D 2) Given
3) ∠B = ∠E 3) Given
4) ΔABC ≅ ΔDEF 4) By ASA postulate
⇒ AB = DE, BC = EF and AC = DF
Consequently, Q coincides with F.
AB BC CA
---- = ------ = ------
DE EF FA
Since the corresponding angles are equal, we conclude that Δ ABC ~ Δ DEF.
Case( ii ) : AB < DE. Then P lies in DE
In triangles ABC and DPQ,
Statements
Reasons
1) AB = DP 1) By construction
2) ∠A = ∠D 2) Given
3) AC = DQ 3) By construction
4) ΔABC ≅ ΔDPQ 4) By SAS postulate
5) ∠B = ∠DPQ 5) CPCTC
6) ∠B = ∠E 6) Given
7) ∠E = ∠DPQ 7) By transitive property
( from above)
8) PQ || EF 8) If two corresponding angles are congruent then the lines are parallel
9) DP/DE = DQ/DF 9) By basic proportionality theorem
10) AB/DE = BC/EF 10) By construction
11) AB/DE = AC/DF 11) By substitution property
12) Δ ABC ~ Δ DEF 12) By SAS postulate
Case ( iii ): If AB > DE. Then P lies on DE produced.
Proof for this case is same as above case ( ii ).
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