Find angle of isosceles triangle

Question

Let $ABC$ an isosceles triangle ($AB=AC$) and the bisector from $B$ intersects $AC$ at $D$ such that $AD+BD=BC$. Find the angle $A$.

I found this in a book and the proof starts with $A=4a$. Then they apply the theorem of sines:

$$AD+BD =BC \implies \frac{AD+BD}{AB}=\frac{BC}{AB} \implies \frac{\sin(45^\circ-a)+\sin 4a}{\sin(45^\circ+3a)}=\frac{\sin 4a}{\cos 2a}$$

I think I understand the angles, but I don't understand how $\cos 2a$ appears. And then they say it implies $a=25^\circ$ so $A=100^\circ$. But I don't understand how $a=25^\circ$.

Please help me understand. Thank you in advance.

Answer 1

Note that, according to the sine rule applies to the triangle ABC,

$$\frac{BC}{AB} = \frac{\sin A }{\sin C} = \frac{\sin 4a }{\sin \frac{180-4a}2}= \frac{\sin 4a }{\sin (90-2a)}=\frac{\sin 4a }{\cos 2a}$$

Answer 2

First, let's see the angles

• $\measuredangle ABD = \dfrac{180^\circ-4a}{4}=45^\circ-a$
• $\measuredangle ADB = 180^\circ-4a-(45^\circ-a)=135^\circ-3a$
• $\measuredangle BDC = 180^\circ-(135^\circ-3a)=45^\circ+3a$
• $\measuredangle C= 90^\circ-2a$

From the theorem of sines in $\Delta ABD$:

$$\frac{AD+BD}{AB}=\frac{\sin(\measuredangle ABD)+\sin(\measuredangle A)}{\sin(\measuredangle ADB)}=\frac{\sin(45^\circ-a)+\sin 4a}{\sin(45^\circ+3a)}$$

and from the theorem of sines in $\Delta ABC$:

$$\frac{BC}{AB}=\frac{\sin (\measuredangle A)}{\sin (\measuredangle C)}=\frac{\sin 4a}{\sin (90^\circ-2a)}=\frac{\sin 4a}{\cos 2a}=\frac{2\sin 2a\cos 2a}{\cos 2a}=2\sin 2a$$

because the $\cos$ function is dephased from $\sin$ by $90^\circ$. Now from the hypothesis:

$$\frac{\sin(45^\circ-a)+\sin 4a}{\sin(45^\circ+3a)}=2\sin 2a$$

and this implies

$$\sin 4a+\sin\left(45^{\circ}-a\right)=2\sin 2a\sin\left(45^{\circ}+3a\right)$$

or

$$\sin 4a+\cos\left(45^{\circ}+a\right)=\cos\left(45^{\circ}+a\right)-\cos\left(45^{\circ}+5a\right)$$

or

$$\sin 4a=\sin \left(5a-45^{\circ}\right)$$

and thus $4a+5a-45^{\circ}=180^\circ\Rightarrow a=25^\circ$. Which means $\measuredangle A=100^\circ$.

Answer 3

Let $E$ be placed on $BC$ such that $EC=AD$.

Thus, $$BD=BC-AD=BC-EC=BE,$$ which says $$\measuredangle BEC=\frac{1}{2}(180^{\circ}-\measuredangle DBC)=\frac{1}{2}(180^{\circ}-45^{\circ}+\alpha)=67.5^{\circ}+\frac{1}{2}\alpha.$$ In another hand, since $BD$ is a bisector of $\Delta ABC$, we obtain: $$\frac{EC}{AB}=\frac{AD}{AB}=\frac{DC}{BC}$$ and $\angle C$ is a common angle of $\Delta ABC$ and $\Delta EDC.$

Thus, $\Delta ABC\sim\Delta EDC,$ which gives $\measuredangle DEC=4\alpha$.

Id est, $$67.5^{\circ}+\frac{1}{2}\alpha+4\alpha=180^{\circ},$$ which gives $\alpha=25^{\circ}$ and $$\measuredangle BAC=4\cdot25^{\circ}=100^{\circ}.$$