Question

In an isosceles triangle ABC, AB = AC, ZBAC = 108° and AD trisects <BAC (divides in
BD
2:1) and BD > DC. The ratio is (Point D lies on side BC)
DC​

Answer 1

Given,

AB=AC, ∠BAC=108

∘

So ∠ABC=∠ACB=

2

180−∠BA

=

2

180−108

=36

∘

In ΔABD

sin36

BD

=

sin108

AB

−(1)

In Δ ACD

sin72

CD

=

sin72

AC

⟹CD=AC

⟹CD=AC=AB

eqs.(1)

⟹

sin36

BD

=

sin108

CD

⟹

CD

BD

=

sin108

sin36

CD

BD

=

3

2