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We know that $\widehat{ACB}=75^\circ$ and that $\left(AB\right)//\left(CD\right)$. We know that $\widehat{CDB}=35^\circ$, and $A, B, C, D$ are on a circle $C$, wich has for center $O$ (not

on the drawing).

I'm trying to get the mesure of $\widehat{DCB}$.

That's what I tried :

$\widehat{ACB}=75^\circ$ and $\widehat{ACB}=\widehat{BDA}$, so $\widehat{BDA}=75^\circ$.

Also,

$\widehat{CDB}=35^\circ$ and $\widehat{CDB}=\widehat{BAC}$, so $\widehat{BAC}=35^\circ$.

But then, I'm blocked, I don't know what else I don't know... How can I use the point $O$ if I don't know where it is on the drawing ? I should I do then ?

Thanks.

Hippalectryon
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Ramanewbie
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  • Fix the formatting to get rid of the "\widehat"s and maybe insert missing equal signs, if they were meant to be there. Otherwise it's hard to read your question. – coffeemath Jan 27 '15 at 19:36

1 Answers1

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$\angle CDA = \angle CDB + \angle BDA = 35^\circ + 75^\circ = 110^\circ$ and $\angle DBA = 35^\circ$ because $CD \parallel BA.$ in a cyclic quad $\angle CBA + \angle CDA = 180^\circ$ so $\angle CBA = 70^\circ$ and using $\angle DBA = 35^\circ$ we have $\angle CBD = 35^\circ$ look at the isosceles triangle $BCD$ and conclude that $\angle BCD = 110^\circ$

abel
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