Ex 6.4, 3
In figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that (ππ(π΄π΅πΆ))/(ππ(π·π΅πΆ)) = π΄π/π·π
Given: β ABC and β DBC
Having common base BC
To prove: (ππ β π΄π΅πΆ)/(ππ β π·π΅πΆ)=π΄π/π·π
Proof:
Since AO and OD are not part of β π΄π΅πΆ and β π·π΅πΆ ,
we cannot directly use the theorem
We know that
Area of triangle = 1/2Γπ΅ππ πΓπ΄ππ‘ππ‘π’ππ
Lets draw altitude AE [ AE β₯π΅πΆ ]
Hence,
ar β ABC=1/2Γπ΅πΆΓπ΄πΈ
Similarly, for Ξ DBC
Let us draw altitude DF (DF β₯ BC)
Hence,
ar β DBC=1/2Γπ΅πΆΓπ·πΉ
Now, taking ratio
(ππ βπ΄π΅πΆ)/(ππ βπ·π΅πΆ) = (1/2 Γ π΅πΆ Γ π΄πΈ)/(1/2 Γ π΅πΆ Γ π·πΉ)
(ππ βπ΄π΅πΆ)/(ππ βπ·π΅πΆ) = π΄πΈ/π·πΉ
Now
in β AOE and Ξ DOF
β AEO = β DFO
β AOE = β DOF
So, by using AA similarity criterion
β AOE ~ β DOF
If two triangle are similar their ,
corresponding sides are in the same ratio
So, π΄πΈ/π·πΉ=π΄π/π·π
Putting (2) in (1)
(ππππ ππ β π΄π΅πΆ)/(ππππ ππ β π·π΅πΆ)=π΄πΈ/π·πΉ
(ππππ ππ β π΄π΅πΆ)/(ππππ ππ β π·π΅πΆ)=π΄π/π·π
Hence proved

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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.