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Given: D and E are mid points of AB and AC
By mid point theorem, DE∥BC
In △ADE and △ABC
∠DAE=∠BAC (Common)
∠ADE=∠ABC (Corresponding angles)
∠AED=∠ACB (Corresponding angles)
Thus, △ABC∼△ADE (AAA rule)
Hence,
A(△ABC)
A(△ADE)
=
AB
2
AD
2
(Similar triangle property)
A(△ABC)
A(△ADE)
=
(2AD)
2
AD
2
A(△ABC)
A(△ADE)
=1:4
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