Question

in triangle ABC seg DE || seg BC. If 2A(triangle ADE)= A(quadrilateral DBCE), find AB: AD and show that BC=√3×DE​

Answer 1

Step-by-step explanation:

3A(△ADE)=A(DECB)⟹A(△ABC)=4A(△ADE)

⟹

A(△ADE)

A(△ABC)

=4

In △ADE and △ABC, ∠DAE=∠BAC

DE∥BC, ∠ADE=∠ABC and ∠AED=∠ACB

⟹△ABC and △ADE are similar.

⟹

A(△ADE)

A(△ABC)

=(

DE

BC

)

2

⟹4=(

DE

BC

)

2

⟹

DE

BC

=

1

2

Answer 2

Step-by-step explanation:

This is ur answer just try to make a diagram first diagrams and required appropriate theorems are like heart and soul of geometry this was not challenging sum pretty easy it was