THEOREM 6.2 Converse Of Proportionality Theorem: If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side. If AD / DB = AE / EC. then DE || BC
Given : A Δ ABC and a line intersecting AB in D and AC in E, such that AD / DB = AE / EC. Prove that : DE || BC Let DE is not parallel to BC. Then there must be another line that is parallel to BC. Let DF || BC. Statements Reasons 1) DF || BC 1) By assumption 2) AD / DB = AF / FC 2) By Basic Proportionality theorem 3) AD / DB = AE /EC 3) Given 4) AF / FC = AE / EC 4) By transitivity (from 2 and 3) 5) (AF/FC) + 1 = (AE/EC) + 1 5) Adding 1 to both side 6) (AF + FC )/FC = (AE + EC)/EC 6) By simplifying 7) AC /FC = AC / EC 7) AC = AF + FC and AC = AE + EC 8) FC = EC 8) As the numerator are same so denominators are equal This is possible when F and E are same. So DF is the line DE itself. ∴ DF || BC
THEOREM 6.3 If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This is also called AAA (Angle-Angle-Angle) criterion. If in two triangles, corresponding angles are equal, then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar. This is also called AAA (Angle-Angle-Angle) criterion.
THEOREM 6.4 If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. If in two triangles, sides of one triangle are proportional to the sides of the other triangle, then their corresponding angles are equal and hence the two triangles are similar. This is also called SSS (Side-Side-Side) criterion. This is also called SSS criteria(side-side-side)
THEOREM 6.5 If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar. This is also called SAS (Side-Angle-Side) criterion.
THEOREM 6.6
= 2 = 2 = 2
Theorem 6.7 If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other .
Theorem 6.8
THEOREM 6.9
Now in triangle PQR , we have PR 2 = PQ 2 +QR 2 (Pythagoras Theorem as ) PR 2 = AB 2 + BC 2 (By construction) (1) But AC 2 = AB 2 +BC 2 (Given) (2) So, AC=PR [From (1)and (2)] (3) Now In triangles ABC and PQR, AB = PQ (By construction) BC = QR (By construction) AC = PR (Proved above in 3) ABC PQR (SSS congruence ) (By CPCT) =90