Please see the attachment. Could anyone tell me how to determine the value of angle x? Thanks!
Hi,
Your problem does have a solution. Unfortunately, I can't figure out how to do it geometrically. The following uses trigonometry to find x. The answer turns out to be very close to 16 degrees. This makes me think there is some geometric solution, but I am at a loss as to what it might be.
Construct EK perpendicular to BC, K on BC and as in your previous post a=BD=DE=EC
$\displaystyle \sin \left(82^{\circ }\right)=\text{EK}/a$
$\displaystyle \sin \left(38^{\circ }\right)=\text{IK}/a$
$\displaystyle \sin (\text{angleEDI})=\text{EI}/a=(\text{EK}-\text{IK})/a=\sin \left(82^{\circ }\right)-\sin \left(38^{\circ }\right)=\sin \left(22^{\circ }\right)$
Proof:
Extend side BA with AF = AC so that angle BFC = 30.
Show that triangles BCD and BFC are similar. Since they have an angle in common, this amounts to showing that BC*BC=BD*BF
This in turn follows from
$\displaystyle \text{DE}^2=\text{AD}^2+\text{AE}^2-\text{AD}.\text{AE}$
and
$\displaystyle \text{BC}^2=\text{AB}^2+\text{AC}^2-\text{AB}.\text{AC}$
(use Pythagoras' theorem)
Therefore angle BCD = angle BFC = 30 degrees