Let $ABCD$ be a convex quadrilateral with these properties:
$\angle ADC = 135$ and $\angle ADB- \angle ABD = 2 \angle DAB = 4 \angle CBD$. If $BC$ $=$ $ \sqrt{2} CD$
prove that, $AB = BC + AD$.
IGO 2016 Elementary/5
- Thamim Zahin
- Posts:98
- Joined:Wed Aug 03, 2016 5:42 pm
I think we judge talent wrong. What do we see as talent? I think I have made the same mistake myself. We judge talent by the trophies on their showcases, the flamboyance the supremacy. We don't see things like determination, courage, discipline, temperament.
Re: IGO 2016 Elementary/5
$\angle DAB +\angle ABD +\angle ADB =180^\circ
\Rightarrow 2\angle CBD + \angle ABD + 4\angle CBD +\angle ABD =180^\circ $
So,$\angle CBD +\angle ABD +2\angle CBD = 90^\circ
\Rightarrow \angle ABC +\angle DAB =90^\circ $
So, $BC \perp AD$ .$E$ be the reflection of $D$ W.R.T $BC$ .Then $ E \in AD$.
$ \angle EBD = 2\angle CBD = \angle DAB$ . So , $\triangle EBD \sim \triangle EAB $ . As $BD = EB \Rightarrow AB = AE =AD + DE = AD + \sqrt{2} CD =AD +BC $ .
\Rightarrow 2\angle CBD + \angle ABD + 4\angle CBD +\angle ABD =180^\circ $
So,$\angle CBD +\angle ABD +2\angle CBD = 90^\circ
\Rightarrow \angle ABC +\angle DAB =90^\circ $
So, $BC \perp AD$ .$E$ be the reflection of $D$ W.R.T $BC$ .Then $ E \in AD$.
$ \angle EBD = 2\angle CBD = \angle DAB$ . So , $\triangle EBD \sim \triangle EAB $ . As $BD = EB \Rightarrow AB = AE =AD + DE = AD + \sqrt{2} CD =AD +BC $ .
The first principle is that you must not fool yourself and you are the easiest person to fool.