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Home » 3. In fig. 4.58, Δ ABC is a triangle such that AB/AC = BD/DC,         find ∠BAD.
3. In fig. 4.58, Δ ABC is a triangle such that AB/AC = BD/DC,

    \[\angle \mathbf{B}=\mathbf{7}{{\mathbf{0}}^{\mathbf{o}}},\]

    \[\angle \mathbf{C}\text{ }=\text{ }\mathbf{5}{{\mathbf{0}}^{\mathbf{o}}},\]

find ∠BAD.
CBSE | Class 10 | Exercise 4.3 | Maths | RD Sharma | Triangles
3. In fig. 4.58, Δ ABC is a triangle such that AB/AC = BD/DC,

    \[\angle \mathbf{B}=\mathbf{7}{{\mathbf{0}}^{\mathbf{o}}},\]

    \[\angle \mathbf{C}\text{ }=\text{ }\mathbf{5}{{\mathbf{0}}^{\mathbf{o}}},\]

find ∠BAD.

Solution:

Given:

 Δ ABCsuch that AB/AC = BD/DC, 

    \[\angle B\text{ }=\text{ }70\]

 o 

 

    \[\angle C\text{ }=\text{ }50\]

o

Required to find: ∠BAD

In ΔABC,

    \[\angle A~=\text{ }180\text{ }\text{ }\left( 70\text{ }+\text{ }50 \right)\]

[Angle sum property of a triangle]

    \[=\text{ }180\text{ }\text{ }120\]

    \[=\text{ }60\]

\frac{AB}{AC}=\frac{BD}{DC},

AD is the angle bisector of angle ∠A.

    \[\angle BAD\text{ }=\angle A/2\text{ }=\text{ }60/2\text{ }=\text{ }{{30}^{o}}\]

i

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