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ABC is an isosceles triangle having $\angle B=\angle C=2*\angle A$. If BD bisecting $\angle B$ meets AC in D,prove that AD=BC.

I know congruent triangles would help but am not able to figure out how to use them. ADB can not be congruent to CBD. I am trying to figure out which triangles might be congruent(triangles with AD as side and BC as side).

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See angles are $72,72,36$ of triangle ABC . So BD is bisector of B implies each angle=36. So $AD=BD ...(1)$isoceles triangle theorem now in triangle BDC angle $BDC=72$ but also $DCB =72$ so $BD=BC$..(2) isoceles triangle theorem thus from 1,2 $AD=BC$ thas all.

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Instead of looking for congruent triangles, you should instead find the angle of triangles themselves.

We know that : \begin{equation} 2\times \angle A=\angle B =\angle C\end{equation}

Since $\angle A, \angle B , \angle C $ are parts of single triangle then : \begin{equation} \angle A+\angle B +\angle C=\pi\end{equation} solve for these angles we would find: \begin{equation} \angle A= \frac{\pi}{5}, \angle B =\angle C=\frac{2\pi}{5}\end{equation} $BD$ bisect $\angle B$ means that

\begin{equation} \angle CBD = \angle ABD =\frac{\pi}{5}\end{equation} Which tells us that triangle $ABD$ is isosceles thus \begin{equation} AD =BD\end{equation}

We can also solve for $\angle BDC$ to find that \begin{equation}\angle BDC=\frac{2\pi}{5}\end{equation} thus triangle $BDC$ is also isosceles and \begin{equation}BD=BC\end{equation} finally we proved that \begin{equation}AD=BC\end{equation}

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By construction we have:

$$ \angle DBA=\angle DCB=\frac{1}{2} \angle ABC=\frac{1}{2} \angle ACB=\angle CAB $$

so, for the triangle $ADB$: $$ \angle DAB= \angle CAB=\frac{1}{2} \angle ABC=\angle DBA \Rightarrow DA=DB $$ and for the triangle $CDB$: $$ \angle CDB=180°-(\angle DCB+\angle DBC)=(180°-\angle DBC)-\angle DCB= $$ $$ =(180°-\angle CAB)-\angle DCB=2\times \angle DCB-\angle DCB=\angle DCB \Rightarrow DB=CB $$

So: $DA=DB$

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