AAA-Similarity
Theorem: If in two triangles, corresponding angles are equal, then their corresponding sides are proportional and the triangles are similar Given: Two triangles, Triangle ABC and A D, B E and C F To prove: Triangle ABC similar to Triangle DEF Proof: Three cases arises Case-I When AB=DE
Triangle DEF such
that
A D, B E and C F and AB=DE, then Triangle ABC Triangle DEF
(ASA property)
Therefore AB DE , BC EF and AC DF (In congruent triangles equal angles have equal sides opposite them)
Therefore
AB BC AC DE EF DF
Thus Triangle ABC similar to Triangle DEF Case II: When ABDE
In this case, mark points P and Q on AB and AC respectively, such that AP=DF and AQ=DF. Join PQ. Now in triangle APQ and DEF AP=DF, AQ=DF and A D
Therefore Triangle APQ Triangle DEF
(SAS property)
So, APQ E
But E B (given)
Therefore APQ B
This implies that PQ parallel to BC
Therefore AP AQ AB AC
(Corresponding angels are equal)
(Corollary to basic proportionality theorem)
So,
DE DF AB AC
… (1)
From (1) and (2), we get
DE DF EF AB AC BC
Triangle ABC similar to Triangle DEF
Corollary: (AA-Similarity). If two angles of a triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
In triangle Triangle ABC and Triangle DEF , if A D, B E then the remaining C of Triangle ABC =remaining F of Triangle DEF . Hence by AAA similarity, Triangle ABC similar to Triangle DEF .
SSS-similarity
Theorem: If the corresponding sides of two triangles are proportional, then their corresponding angles are equal and the triangles are similar. Given: Two triangles Triangle ABC and Triangle DEF such that To prove: Triangle ABC similar to Triangle DEF Construction: Mark points P and Q on DE and DF respectively such ath DP=AB and DQ=AC. Join PQ.
AB BC AC DE EF DF
Proof: Since
AB AC and AB=DP, AC=DQ DE DF
Therefore
DP DQ . This implies that PQ parallel to EF DE DF
(Converse of Basic proportionality theorem) So, DPQ E and DQP F (corresponding angles are equal)
Triangle DPQ similar to Triangle DEF
(By AA similarity)
… (1)
So,
DP PQ DE EF AB PQ DE EF AB BC DE EF
or
( Since DP AB )
But
BC PQ EF EF
This implies that BC=PQ Now AB=DP, AC=DQ and BC=PQ This implies that Triangle ABC Triangle DPQ (SSS congruency) … (2)
From (1) and (2), we get
Triangle ABC similar to Triangle DEF
SAS-Similarity
Theorem: If one angle of a triangle is equal to one angle of the other and the sides including these angles are proportional, the triangles are similar.
Given:
Two triangles AB AC A D, . DE DF
Triangle ABC and
Triangle DEF such
that
To prove: Triangle ABC similar to Triangle DEF Construction: Mark points P and Q in DE and DE respectively such that DP=AB and DQ=AC. Join PQ. Proof: In Triangle ABC and Triangle DPQ , we have AB=DP, AC=DQ and A D
Therefore Triangle ABC Triangle DPQ (SAS congruency criterion)
… (1)
Now,
AB AC this implies that DE DF
DP DQ DE DF
(AB=DP, AC=DQ) ((converse of Basic proportionality Theorem)
So, PQ parallel to EF
Thus DPQ E and DQP F
(Corresponding angles)
Therefore Triangle DPQ similar to Triangle DEF (AA Similarity Criterion) … (2)
From (1) and (2)
Triangle ABC similar to Triangle DEF