AD is an altitude of an equilateral triangle ABC. On AD as base another equilateral triangle ADE is constructed. Prove that ar(∆MDE): ar(∆ABC)
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Step-by-step explanation:
We have an equilateral △ABC in which AD is altitude. An equilateral △ADE is drawn using AD as base.
Since, the two triangle are equilateral, the two triangles will be similar also.
△ADE∼△ABC
We know that according to the theorem, the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.
⇒
ar(△ABC)
ar(△ADE)
=(
AB
AD
)
2
----- ( 1 )
Now, △ABC is an equilateral triangle.
∴ ∠B=60
o
⇒ sinB=
AB
AD
⇒ sin60
o
=
AB
AD
⇒
2
3
=
AB
AD
⇒ (
AB
AD
)
2
=
4
3
Substituting above value in equation ( 1 ) we get,
⇒
ar(△ABC)
ar(△ADE)
=
4
3
------ Hence proved
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