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Given this sketch: enter image description here

Angles $ABC$ and $BDE$ are right angles and $|BD|=|BC|$. I want to prove that triangle $AED$ is isosceles.

Obviously I used Pythagoras for triangles $ABC$ and $BED$ but so far this didn't get me anywhere.

I'd really appreciate help. Thank you!

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in the triangle $$\Delta BDC$$ we have $$BD=BC=R$$ s0 $$\angle CDB=\angle BCD=\gamma$$ and we get $$\gamma +90^{\circ}+x=180^{\circ}$$ so $$x=90^{\circ}-\gamma$$ but $$\alpha+\gamma=90^{\circ}$$ and we get $$x=90^{\circ}-(90^{\circ}-\alpha)=\alpha$$ therefore $$AE=ED$$

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$\angle BCA=90^\circ-\angle BAC$. At the same time$\angle BCA=\angle BDC$. We know that $\angle BDC+\angle BDE+\angle EDA=180^\circ$ and that $\angle BDE=90^\circ$. Therefore $\angle EDA=90^\circ-\angle BCD=\angle BAC=\angle EAD$

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The trick is to extend CB to cut the circle at C’. Then, CC' is a diameter of the circle with $\angle CDC’ = 90^0$.

enter image description here

$\angle A = 90^0 – x$

Also, $y = … = z = 90^0 - t$

Result follows because $x = t$.

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  • $\begingroup$ You proved that the BCD triangle is isosceles (which was already obvious, because BC and BD are equal, being rays of the same circle). However, the requirement was to prove that AED triangle is isosceles. $\endgroup$ – Razvan Socol Feb 18 '18 at 18:20
  • $\begingroup$ @RazvanSocol If we reverse the whole process, then from $x = t$, we have $90^0 - x = 90^0 - t$. This means $\angle A = y$. Therefore, .... $\endgroup$ – Mick Feb 18 '18 at 18:26
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Add a tangent at C. Then you have a new triangle formed by the secant CD, the tangent through C, and the tangent through D. It is easy to show this is isosceles. And it is similar to triangle AED since all legs are parallel.

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