In a given diagram equilateral triangles triangle abc and triangle ecd are standing on the same straight line bcd.prove that a) angle acd= angle ecb b) ad = be
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Given that:- △ABC is an isosceles triangle and ∠ABC=90°
∴AB=BC
△ABE∼△ACD(∵All equilateral triangles are similar)
To find:-
ar(△ACD)
ar(△ABE)
=?
Solution:-
In △ABC,
Using pythagoras theorem,
AC
2
=AB
2
+BC
2
AC
2
=AB
2
+AB
2
[∵AB=AC]
AC
2
=2AB
2
.....(i)
Now In △ABE and △ACD
△ABE∼△ACD(Given),
As we know that ratio of area of similar triangles is equal to the ratio of squares of their corresponding sides.
∴
ar(△ACD)
ar(△ABE)
=
AC
2
AB
2
⇒
ar(△ACD)
ar(△ABE)
=
2AB
2
AB
2
[From(i)]
⇒
ar(△ACD)
ar(△ABE)
=
2
1
⇒ar(△ABE):ar(△ACD)=1:2
Hence the ratio between the area of △ABE to the area of △ACD is 1:2.
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