Theorem 6.4 (SSS Criteria) : If in two triangles, sides of one triangle are proportional to (i.e., the same ratio of) the sides of the other triangle, then their corresponding angles are equal and hence the two triangle are similar. Given: Triangle βABC and βDEF
such that π΄π΅/π·πΈ = π΅πΆ/πΈπΉ = πΆπ΄/πΉπ·
To Prove: β A = β D, β B = β E, β C = β F
and βABC ~ βDEF
Construction: Draw P and Q on DE & DF
such that DP = AB and DQ = AC respectively
and join PQ.
Proof:
Given
π΄π΅/π·πΈ = πΆπ΄/π·πΉ
And DP = AB, DQ = AC
π·π/π·πΈ = π·π/π·πΉ
π·πΈ/π·π = π·πΉ/π·π
Subtracting 1 on both sides
π·πΈ/π·π β 1 = π·πΉ/π·π β 1
(π·πΈ β π·π)/π·π = (π·πΉ β π·π)/π·π
ππΈ/π·π = ππΉ/π·π
π·π/ππΈ = π·π/ππΉ
Using Theorem 6.2 : If a line divides any two sides of a triangle in the same ratio, then the line is parallel to third side.
β΄ PQ β₯ EF.
Now,
For lines PQ & EF,
with transversal PE
β P = β E
For lines PQ & EF,
with transversal QF
β Q = β F
In βDPQ and βDEF
β P = β E
β Q = β F
β βDPQ ~ βDEF
β΄ π·π/π·πΈ = π·π/π·πΉ = ππ/πΈπΉ
Also,
π΄π΅/π·πΈ = π΅πΆ/πΈπΉ = πΆπ΄/πΉπ·
& AB = DP & CA = QD
Thus, π·π/π·πΈ = π΅πΆ/πΈπΉ = π·π/πΉπ·
From (2) and (3)
π΅πΆ/πΈπΉ = ππ/πΈπΉ
β΄ BC = PQ
In βABC and βDPQ
AB = DP
AC = DQ
BC = PQ
β βABC β βDPQ
β΄ β B = β P
β C = β Q
β A = β D
But From (1)
β P = β E and β Q = β F
Therefore,
β B = β P = β E
and β C = β Q = β F
Therefore,
In Ξ ABC & Ξ DEF
β B = β E
β C = β F
β΄ βABC ~ βDEF
Hence Proved

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.