# find angle ACB

#### Albert

##### Well-known member
$\triangle ABC , \angle ABC=45^o,\,\, point \,\, D\,\, on \,\, \,\overline{BC}$

$and,\,\, 2\overline{BD}=\overline{CD},\,\, \angle DAB=15^o$

$find :\,\, \angle ACB=?$

#### caffeinemachine

##### Well-known member
MHB Math Scholar
$\triangle ABC , \angle ABC=45^o,\,\, point \,\, D\,\, on \,\, \,\overline{BC}$

$and,\,\, 2\overline{BD}=\overline{CD},\,\, \angle DAB=15^o$

$find :\,\, \angle ACB=?$
Find a point $M$ between $A$ and $D$ such that $\angle MBD=30$. Note that $MD=DM$ and $AM=MB$. Say $BD=1$. The above leads to $AD=\sqrt 3 + 1$. Further note that $\angle CMD=90$ and hence $\angle MCA=45$. Consequently $\angle ACB=75$.

#### Albert

##### Well-known member
Find a point $M$ between $A$ and $D$ such that $\angle MBD=30$. Note that $MD=DM$ and $AM=MB$. Say $BD=1$. The above leads to $AD=\sqrt 3 + 1$. Further note that $\angle CMD=90$ and hence $\angle MCA=45$. Consequently $\angle ACB=75$.
caffeinemachine :very good solution

#### caffeinemachine

##### Well-known member
MHB Math Scholar
caffeinemachine :very good solution
Thanks. If you have a different one then please post it.

#### Albert

##### Well-known member

from the diagram it is easy to see that :
$\angle ACB =30^o +45^o =75^o$

MHB Math Scholar