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Let $A,B,C$ be a triangle's vertices (FIG.1), and $D$ is a point on $\overline {BC}$. It is given that $\angle B= 2 \angle C $, $\overline{AB}=\overline{DC}$ and also that $\overline{AD}$ is the bisector of angle $A$.. We have to prove that angle $\angle BAC=72°$.

$\hspace{0.54 cm}$ This is not a triangle. $$\begin{align}\text{FIGURE 1} &\end{align}$$

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Isn't there supposed to be a figure included? –  J. M. Aug 12 '12 at 5:45
    
...or at least let us hope he'll tell us what are A,B,C (a triangle's vertices?), and...D a point on BC? Gee, some info missing here! –  DonAntonio Aug 12 '12 at 6:01
    
@DonAntonio There is no information missing. The question is good as it stands. A figure would help, of course. I'll supply one. –  Pedro Tamaroff Aug 12 '12 at 6:24
    
@J.M. There you go. –  Pedro Tamaroff Aug 12 '12 at 6:43
    
@PeterTamaroff, of course there's info missing, and a lot. The thing is we're very good guessers, but that question is very poorly stated –  DonAntonio Aug 12 '12 at 6:48
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2 Answers

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Let $BE$ bisect $\angle B$ into a pair of angles congruent to $\angle C$. As $\angle CBE$ is a member of that pair, we have an isosceles triangle with $BE=EC$ (blue edges).

enter image description here

Draw $DE$ and note that, because $\angle ABE = \angle C$, we have $\triangle ABE \cong \triangle DCE$ ("side-angle-side"). In particular, $\angle CDE \cong \angle A$ (red-green angles).

enter image description here

Also, $DE\cong EA$ (green edges), so that $\triangle DEA$ is isosceles with $\angle EAD \cong \angle EDA$. As $\angle A$ was bisected by $AD$, we have $\angle EDA \cong \angle BAD$, so that $DE \parallel BA$ ("alternate interior angles", or what I call the "Z theorem").

enter image description here

Consequently, $\angle CDE \cong \angle CBA$ ("corresponding angles", or the "F theorem"), which implies that all of my angle-marking dots match. With five dots in the angles of $\triangle ABC$, each dot represents measure $180^{\circ}/5=36^{\circ}$, so that the two-dot $\angle A$ has measure $72^{\circ}$. QED

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As $B=2C \Rightarrow A=\pi-B-C=\pi-3C$ then

$$ 0<B<\pi \Rightarrow 0<2C<\pi \Rightarrow 0<C<\frac{\pi}{2}$$

and $A=\pi-3C$ so, $0<A<\pi \Rightarrow0<\pi - 3C <\pi \Rightarrow0< C< \frac{\pi}{3} $

So, $0< C< \frac{\pi}{3} $

If $∠BAD=x , ∠DAC=x$ and if $∠ADB=y, ∠ADC=\pi-y$

If $AB=c$, $BC=a$, $CA=b \Rightarrow BD=a-c$

Applying the law of sines on triangle $ACD$, $$\frac{c}{\sin x} = \frac{b}{\sin (\pi- y)}\Rightarrow\frac{c}{\sin x} = \frac{b}{\sin y}$$

Applying sine law on triangle ABD, $$\frac{a-c}{\sin x} = \frac{c}{\sin y}$$

Upon division, $$\frac{a-c}{c}=\frac{c}{b}$$

$$\frac{\sin\ A - \sin C}{\sin C}=\frac{\sin C}{\sin B}$$ (applying $a=2R\sin A$ etc.)

$$\frac{\sin\ (\pi - 3C) - \sin C}{\sin C}=\frac{\sin C}{\sin 2C}$$

$$\frac{\sin\ 3C - \sin C}{\sin C}=\frac{\sin C}{\sin 2C}$$

$$(2\sin C \cos2C)\sin2C = (\sin C)^2$$

applying $$\sin2x-\sin2y=2\sin(x-y)\cos(x+y)$$

$$\sin C(2 \cos2C\sin2C - \sin C)=0$$

$$\sin C(\sin 4C - \sin C)=0$$

$$\sin 4C=\sin C \text{ as } 0<C<\pi=>\sin C>0$$

$4C=n\pi+(-1)^nC$ where n is any integer,

If $n$ is even $(=2m)$,say, $4C=2m\pi+C \Rightarrow C=\frac{2m\pi}{3}$,

As $0< C< \frac{\pi}{3} $, so there is no solution if $n$ is even.

If $n$ is odd $(=2m+1)$, say, $4C=(2m+1)\pi-C\Rightarrow C=\frac{(2m+1)\pi}{5}$

$$C=\frac{\pi}{5}\ as\ 0< C< \frac{\pi}{3} $$

$$A=\pi-3C=\frac{2\pi}{5}=72^\circ$$

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