<CAD=? AB//CD, <DBC=10^o <BCD=130^o, AB=AD 2016 Armenia NMO 8.6

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<CAD=? AB//CD, <DBC=10^o <BCD=130^o, AB=AD 2016 Armenia NMO 8.6

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#1 h Aug 18, 2021, 5:36 PM

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In quadrilateral $ABCD$ , $AB\parallel CD$ , $\angle DBC = 10^o$ , $\angle BCD=130^o$ and $AB = AD$ . Find the measure of the angle $\angle CAD$ .

This post has been edited 2 times. Last edited by parmenides51, Aug 18, 2021, 6:20 PM

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#2 h Aug 18, 2021, 11:23 PM

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I am getting that this is not possible? We have $\angle BDC = 40$ , so $\angle ABD = 40$ from parallel lines. But $\angle DAB = 180 - \angle BCD = 50$ , so this means $50 + 40 \cdot 2 = 180$ which is obviously wrong.

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#3 h Aug 18, 2021, 11:42 PM • 2 Y

Y by centslordm, parmenides51

Why is $\angle DAB = 180^\circ - \angle BCD$ ?

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#5 h Aug 19, 2021, 3:07 PM

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mmmm this is what I have so far and I'm really frustrated right now because I feel really close to the answer but also really far...

This post has been edited 1 time. Last edited by Equinox8, Aug 19, 2021, 3:46 PM

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#6 h Aug 19, 2021, 3:28 PM

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datoucai wrote:

Why is $\angle DAB = 180^\circ - \angle BCD$ ?

Because it is a trapezoid?

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#7 h Aug 19, 2021, 3:29 PM

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@above
no, have a look again, the correct is $\angle DAB = 180^\circ - \angle ADC$ since $AB\parallel CD$

the result is

This post has been edited 3 times. Last edited by parmenides51, Aug 19, 2021, 3:33 PM

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#9 h Aug 19, 2021, 3:34 PM

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Nevermind, I was not being smart.

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#10 h Aug 19, 2021, 3:43 PM

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Okay, now I let $\angle ACD = x$ and chased all angles, but then what do I do?

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#11 h Aug 19, 2021, 3:46 PM

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OlympusHero wrote:

Okay, now I let $\angle ACD = x$ and chased all angles, but then what do I do?

Personally I am stuck making any progress other than finding values for all the angles in terms of $x$ .

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AlanLG

#12 h Aug 19, 2021, 5:07 PM

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Let $\angle ACD=x$ , it´s easy to obtain, $\angle DAC=100-x, \angle ABC=50, \angle BCA=130-x$

By law of sines in

$\triangle ABC: \frac{AB}{AC}=\frac{\sin (130-x)}{\sin50}$

$\triangle ACD: \frac{AD}{AC}=\frac{\sin x}{\sin 80}$

As, $AB=AD$

$\frac{\sin (130-x)}{\sin 50}=\frac{\sin x}{\sin 80}$

$x=80$ clearly satisfies, thus $100-x=\angle CAD=20^\circ$

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UnprovableTruth

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UnprovableTruth

#13 h Aug 19, 2021, 5:44 PM

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I think I found another solution if anyone is interested, though it probably isn't nearly the simplest one: Click to reveal hidden text

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Desimathematics

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Desimathematics

#14 h Aug 19, 2021, 6:33 PM

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another solution

Draw a line parallel to $AD$ through b that meets $CD$ in E. Now we prove that $ABEC$ is a cyclic quadrilateral

PROOF
1) we can see that ABDE is a rhombus.
2)the diagonal $AE$ bisects the angles $DAB$ and $BED$ therefore $\angle BAE = 50$
3) Also $\angle BCE = 50 = \angle DBC +\angle BDC$
4) this proves that $ABCE$ is cyclic

now , we have $\angle CAE = \angle CBE = 30$ (angles in same segment)
but we know that $\angle DAE = 50$

So , $\angle CAD = 50 - 30 = 20$

This post has been edited 2 times. Last edited by Desimathematics, Aug 19, 2021, 6:42 PM

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#15 h Aug 21, 2021, 4:51 PM

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Solution

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