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In triangle $ABC$, the bisector of angle $A$ meets side $BC$ in point $D $ and the bisector of angle $B$ meets side $AC$ in point $E$. Given that $DE$ is parallel to $AB$, show that $AE = BD$ and that the triangle $ABC$ is isosceles.


The solution of this problem is completely based on pure geometry. Using alternate angles and basic proportionality theorem. I however want to solve the problem using vectors. We know $$\vec {AB}=\lambda \vec {ED} $$ also $$\widehat{BE}=\widehat{BD}+\widehat{DA} $$ $$\widehat{AD}=\widehat{AE}+\widehat{AB} $$ I am stuck here.

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    $\begingroup$ Angle bisectors are awkward using vectors. $\endgroup$ – almagest Apr 22 '16 at 15:48
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    $\begingroup$ What are the second and third equations? Do the hats mean unit vectors? $\endgroup$ – coffeemath Apr 22 '16 at 16:30
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Let $\overrightarrow{AB}=\underline{a}$ and $\overrightarrow{AC}=\underline{b}$

Then $$\overrightarrow{AD}=\lambda\left(\frac{\underline{a}}{|a|}+\frac{\underline{b}}{|b|}\right)$$

And likewise, $$\overrightarrow{BE}=\mu\left(-\frac{\underline{a}}{|a|}+\frac{\underline{b}-\underline{a}}{|\underline{b}-\underline{a}|}\right)$$

Also let $\overrightarrow{BD}=p(\underline{b}-\underline{a})$ and $\overrightarrow{AE}=q\underline{b}$

Then $$\overrightarrow{AD}=\underline{a}+p(\underline{b}-\underline{a})=\lambda\left(\frac{\underline{a}}{|a|}+\frac{\underline{b}}{|b|}\right)$$

It therefore follows that $p=\frac{\lambda}{|b|}$ and $1-p=\frac{\lambda}{|a|}$

Therefore solving these gives $$p=\frac{|a|}{|a|+|b|}$$

Following the same routine, we also find that $$q=\frac{|a|}{|\underline{b}-\underline{a}|+|a|}$$

Then $$\overrightarrow{DE}==\frac{|a|\underline{b}}{|\underline{b}-\underline{a}|+|a|}-\underline{a}-\frac{|a|(\underline{b}-\underline{a})}{|a|+|b|}$$

This vector is parallel to $\underline{a}$ so it follows that $$\frac{|a|}{|\underline{b}-\underline{a}|+|a|}=\frac{|a|}{|a|+|b|}$$

Thus it follows that $|\underline{b}-\underline{a}|=|b|$ and the triangle is isosceles. We can also deduce that $|AE|=|BD|$.

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Observe that for DE || AB and AD as transversal, alternate angles ADE and DAB are equal. And AD being the bisector of angle BAC, angles DAB and DAC are equal. So we have angles DAC and ADE equal and in triangle ADE, we have angles DAE and ADE equal. So we can say that DE=AE.

Similarly, we again observe that for DE || AB and BE as transversal, alternate angles BED and EBA are equal. And BE being the bisector of angle ABC, angles EBA and EBC are equal. So we have angles BED and EBC equal and in triangle ADE, we have angles BED and EBD equal. So we can say that DE=BD.

Comparing above results, we conclude that AE=BD.

Can you use this result and prove the second case? It is easy. Try it.

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  • $\begingroup$ I specially want to solve the problem using vectors. $\endgroup$ – mathemather Apr 22 '16 at 15:57

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