Internal Bisector of a triangle

Question

In $\Delta ABC$, the bisector of the angle A meets the side BC at D and circumscribed circle at E, then DE equals to

(A) $\frac{{{a^2}\cos \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(B) $\frac{{{a^2}\sec \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(C) $\frac{{{a^2}\sin \frac{A}{2}}}{{2\left( {b + c} \right)}}$

(D) $\frac{{{a^2}\cos ec\frac{A}{2}}}{{2\left( {b + c} \right)}}$

My approach is as follow

Internal Bisector property

$\frac{{AB}}{{AC}} = \frac{{BC}}{{DC}} \Rightarrow \frac{c}{b} = \frac{t}{{a - c}}$

How do I proceed in order to find DE

Answer 1

By Sine Theorem Since $\angle ABC=\angle AEC$ and $\angle ACB=\angle AEB$ (angles are equal subtended same segment)

$$\frac{a}{\sin\angle A}=\frac{b}{\sin\angle AEC}=\frac{c}{\sin\angle AEB}=\frac{b+c}{\sin\angle AEC + \sin \angle AEC}$$

$$\sin\angle AEC + \sin \angle AEC=\frac{(b+c)\sin\angle A}{a}$$

$\angle BCE=\angle CBE=\angle BAE=\angle CAE=\frac{A}{2}$ (angles are equal subtended same segment and $AE$ bisects $\angle A$)

$$\frac{DE}{\sin\angle \frac {A}{2}}=\frac{DC}{\sin\angle AEC}=\frac{BD}{\sin\angle AEB}=\frac{BD+DC}{\sin\angle AEC + \sin \angle AEC}=\frac{a}{\sin\angle AEC + \sin \angle AEC}$$

$$\sin\angle AEC + \sin \angle AEC=\frac{a\sin\angle \frac {A}{2}}{DE}$$

$$\frac{(b+c)\sin\angle A}{a}=\frac{a\sin\angle \frac {A}{2}}{DE}$$

$$DE=\frac{a^2\sin\angle \frac {A}{2}}{2(b+c)\sin\angle \frac {A}{2}\cos\angle \frac {A}{2}}=\frac{a^2}{2(b+c)\cos\angle \frac {A}{2}}$$

Answer 2

as $A,B,C,E$ are concyclic.$$ED.DA=BD.DC$$ using this Formula:$$AD=\frac{2\sqrt{bcs(s-a)}}{b+c}$$ Aslo by angle bisector rule$$BD=\frac{ca}{b+c},CD=\frac{ba}{b+c}$$

Can you finosh it?

Answer 3

$[ABC] = \frac{ab \sin c}2$. Let $D$ be the feet of the internal $A$ bisector.

$[ADB] + [ADC] = [ABC] \iff cd\sin(\frac{\angle A}2) + bd \sin(\frac{\angle A}2) = bc \sin(\angle A) \iff$

$ AD = \frac{2bc \cos (\frac{\angle A}2)}{b+c}$

Now $DE$ seems a bit trickier, how about:

$ce\sin(\frac{\angle A}2) + be \sin(\frac{\angle A}2) = bc \sin(\angle A) + \frac{a^2}2 \tan (\frac{\angle A}2)$ ?

$e = \frac1{b+c} (2bc \cos (\frac{\angle A}2) + \frac{a^2}{2 \cos (\frac{\angle A}2)}) = \frac{\sec (\frac{\angle A}2)}{b+c}(\frac{a^2}2 + bc(\cos (\angle A)+1))$

Answer 4

Since a full derivation is not required, take relative proportions and well known properties of an easily calculable equilateral triangle of side $2, \angle BAO =30^{\circ}, $ altitude is divided by center in ratio $2:1$ etc.

$$ OD=DE= \dfrac{1}{\sqrt 3}$$

Only option (B) gives this value.