Some triangle ... right

Let $\angle{BAC}=x$ and $\angle{DBC}=y$ . Then by Law of Sines we have $\frac{\sin{x}}{BD}=\frac{\sin{65}}{AD}, \frac{\sin{25}}{BD}=\frac{\sin{y}}{CD}$ . Using $AD=CD$ , we get $\sin{x}\sin{y}=\sin{25}\sin{65}$ . Then use product to sum formula to arrive at $\cos{(x-y)}=\cos{40}$ . So $x-y=\pm 40$ and $x+y=90$ . Therefor the possible values for $x$ are 65 and 25. So $\angle{BDC}$ can be either 130 or 90.

There are two answers 130 and 90. If angle ABC=90, then it's very easy to see angle BDC=130.

If angle ABC is not 90, drop an altitude from A to BC, let the foot be X. Then angle AXD=XAD=65=ABD, so A,X,B,D are concyclic. Hence BDC=AXB=90.

It's pretty clear to see an isosceles triangle ABC with angle ABC= 130 would satisfy the condition in the problem.

Law of sines will do the trick

Even u can use m'n theorem

@Calvin Lin I think this problem can be solved without using trignometry.Is I'm right??

This is too easy to be a level 4 I think that everyone should be able to solve it...

Here is a solution to the problem using elementary trigonometry. From the well-known trigonometric identity: (m+n)cot( BDC )=ncot( DBA )-mcot( DBC ) which holds for any triangle ABC and any cevian BD such as CD:AD=m:n , we have, for the above mentioned triangle, the equation: 2cot( BDC )=cot( 65 )-cot( 180 - 25 - BDC ) , which is obtained by setting m=n=1 , and the values of the respective angles (in degrees) as given. Writing the cotangents as the ratios of the cosines and sines, and then by very little trigonometric manipulation, the equation reduces to cos( BDC - 40 )=0=cos 90 *. Since *angle **BDC is less than 180 degrees, this reduced equation can be written in the form: (BDC-40)=90 , from which we get BDC=90+40=130 , which is the required answer.

First off, I would like to point out that this problem has two reasonable solutions: $130°$ and $90°$ . I believe the latter was ignored because would have been too obvious, since it can be easily found if one makes false assumptions, such as that $BD$ is a perpendicular bisector of $AC$ , or that $BD$ bisects $∠ABC$ , etc. In either case, this problem should be fixed so that there is only one definitive solution.

Now for the actual solution. Drawing a diagram, I labeled $AD$ as $x$ and $∠ADB$ as $y$ . Therefore, in terms of these assignments, $∠BAD=180°-(65°+y)=115°-y$ , and $∠DBC=180°-(180°-y+25°)=y-25°$ . Using the Law of Sines on $\Delta ADB$ and $\Delta CDB$ , we see that:

1. $\frac{\sin 65°}{x}=\frac{\sin (115°-y)}{BD}$
2. $\frac{\sin 25°}{BD}=\frac{\sin (y-25°)}{x}$

Solving for BD, we get $BD=\frac{x \sin (115°-y)}{\sin 65°}$ and $BD=\frac{x \sin 25°}{\sin (y-25°)}$ . Setting these equal to each other, the $x$ 's cancel, and our final expression is $\frac{\sin 25°}{\sin (y-25°)}=\frac{\sin (115°-y)}{\sin 65°}$ . Though I haven't been able to rigorously prove it yet, the only two reasonable cases are when both sides are equal to $\frac{\sin 25°}{\sin 65°}$ or when both sides are equal to 1. Using the first case, $\frac{\sin 25°}{\sin (y-25°)}=\frac{\sin 25°}{\sin 65°} \Rightarrow y=90°$ . Using the second case, $\sin 25°=\sin (y-25°) \Rightarrow y=50°$ . Since the angle we are looking for is $180°-y$ , we obtain the two answers: $\boxed{m∠BDC=130°}$ or $m∠BDC=90°$ .