Join Sarthaks eConnect Today - Largest Online Education Community!
+2 votes
222 views
asked in Mathematics by (8k points)
D and E are points on the sides AB and AC respectively of a triangle ABC such that DE is parallel to BC and divides triangle ABC into two parts equal in area. Prove that BD/AB=(2-sq.rt of 2)/2

1 Answer

+3 votes
answered by (21.2k points)
selected by
 
Best answer

Solution:
Given: ABC is a triangle with DE || BC divides the triangle into two parts equal in areas.
To find: BD/AB=(2-sq.rt of 2)/2

Area of ΔADE = Area of Trapezium BDEC (As given)

∴ Area of ΔADE = 1/2 area of ΔABC ------------(1)

In ΔADE and ΔABC

∠ADE = ∠ABC (Corresponding angles)
∠AED = ∠ACB (Corresponding angles)
ΔADE ∼ ΔABC (AA similarity) 

∴ Area of ΔADE / Area of ΔABC = AD2/AB2  (Areas of similar triangle) --------(2)

∴ From equation 1 and 2 we get

=> 1/2 = AD2/AB2

=> AD/AB = 1/√ 2

∴ AB – AD = √ 2 AD – AD
∴ BD = (√ 2 – 1)AD

∴ BD/AB = (√ 2 – 1)AD / √ 2 AD

∴ BD/AB = (√ 2 – 1) / √ 2  --------------(3)

Multiplying Nr and Dr of equation 3 by √ 2 , we get

∴ BD/AB = (2 – √ 2) / 2

commented by (27.7k points)
@Best Answer

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

One Thought Forever

“There is a close connection between getting up in the world and getting up in the morning.“
– Anon
~~~*****~~~

Categories

...