ABC and D

draw the perpendicular line from B to DC and label the intersection as point E

Consider Right Triangle BAE, since Triangle BAE is a 30-60-90 triangle with AB as its hypotenuse, we have BE=(1/2)(AB)

Consider Right Triangle BEC, since Triangle BEC is a 45-45-90 triangle with BC as its hypotenuse, we have BE=EC = (1/2)(AB)

it is given that AB = DC, and DC = DE + EC, we have DE = (1/2)(AB)

consider Rigt Triangle BED with hypotenuse BD, since BE = DE, we have angle EDB and angle DBE to be both 45 degrees

since angle EDB is 45 degrees, we have angle ADB = 180 - angle EDB

angle ADB= 180-45 angle ADB= 135 degrees

Let $H$ be the point on line segment $AC$ such that $\angle AHB = \angle BHC = 90 ^ \circ$ : in order to that $\angle ABH = 60 ^ \circ$ and $\angle HBC = 45 ^ \circ$ .

Since $ABH$ is a right triangle whose angles are $30 ^ \circ$ , $60 ^ \circ$ and $90 ^ \circ$ , $AB=2 \times BH$ and considering that $HBC$ is a right triangle whose angles are $45 ^ \circ$ , $45 ^ \circ$ and $90 ^ \circ$ , therefore $BH=HC$ .

For the reason that $HC= \frac {1}{2} \times AB$ , $H$ is the middle point of the line segment $DC$ .

$DBC$ is a isosceles triangle, because $H$ is the middle point of the base $DC$ and $BH$ is the height of the triangle $DBC$ from the vertex $B$ . In an isosceles triangle, the angles opposite the equal sides are equal, so $\angle BDC = \angle BCD= 45 ^ \circ$ . Hence, $\angle ADB = 180 ^ \circ - \angle BDC = 180 ^ \circ-45 ^ \circ = 135 ^ \circ$

First, quickly check that $D$ is in between $A$ and $C$ - $\angle ABC = 180^\circ - 30^\circ - 45^\circ = 105^\circ$ , which is bigger than $\angle BCA$ . Therefore $AC > AB$ , so when $DC$ is drawn to the same length as $AB$ the point $D$ is in between $A$ and $C$ .

Using the sine rule on the triangle $ABC$ we get that $\frac{AB}{sin(45^\circ)} = \frac{BC}{sin(30^\circ)}$ or, evaluating the sines, $AB = BC \sqrt{2}$ . Therefore $DC = BCsqrt{2}$ .

At this point we look at the triangle $BCD$ . You could notice that it has the pattern of a $45^\circ:45^\circ:90^\circ$ right angled triangle, as the side $DC$ is $\sqrt{2}$ times as big as $BC$ with a $45^\circ$ angle in between. The next section proves this trigonometrically.

Let $x = \angle DBC$ . Then $\angle CDB = 180^\circ - \angle BCD - \angle DBC = 135^\circ - x$ . Using the sine rule with these two angles gives $\frac{DC}{sin(x)} = \frac{BC}{sin(135^\circ - x)}$ . Rearranging, substituting in $DC = BC\sqrt{2}$ and cancelling gives

$sin(x) = sin(135^\circ - x)\sqrt{2}\\ \quad\quad\:\:\:\, = (sin(135^\circ)cos(x) - cos(135^\circ)sin(x))\sqrt{2}\\ \quad\quad\:\:\:\, = (cos(x)\frac{\sqrt{2}}{2} + sin(x)\frac{\sqrt{2}}{2})\sqrt{2}\\ \quad\quad\:\:\:\, = cos(x) + sin(x)$

So $cos(x) = 0$ , which means that $\angle DBC = 90^\circ$ and $DBC$ is a right angled triangle with the ratios mentioned earlier. In particular, $\angle CDB = 45^\circ$ and, since $A$ , $D$ and $C$ are on the same line, $\angle ADB = 180^\circ - \angle CDB = \underline{135^\circ}$ .

Label the foot of the altitude from B as point E, and denote the height as h. Then, triangle BAE is a 30-60-90 triangle with right angle at E, and thus AB = 2h, which also equals DC, as given in the problem. Also, note that triangle BEC is a 45-45-90 triangle, so CE = h. Then, DE = DC - CE = h, which means that BED is another 45-45-90 triangle. Thus, angle ADB measures 180-45 = 135 degrees.

By the Sine Rule on the triangle ABC, we have: BC/sin (30º)= AB/sin (45º) , where AB=2.BC.sin(45º) (*).

Let E be over AC such that BE is the height for B. How the triangle CEB is rectangle, we have EC/BC=sin(45º), where EC=BC.sin(45º)=AB/2 for (*). But, by hypothesis, CD=AB, where EC=CD/2. Then EC=ED=CD/2, that is, BE is height for B (on triangle CBD) and median. Thus CBD is a isosceles triangle with BD=BC. Then (angle)BCD=(angle)BDC=45º and so (angle)BDA=180º-45º=135º.

Notes: "." is the multiplication symbol "/" is the division symbol

There are a few ways you could solve this problem. Of course you could use a ruler and a protractor, but that would be cheating and we want a mathematical way to solve it. I solved this problem like this:

Start off by drawing triangle ABC according to the requirements. I gave AB a set value of 10 since it didn't affect the problem. Using the Law of Sines, AB/sin C=BC/sin A. Plugging in the numbers you end up with 10/≈0.7071067 =BC/0.5. Eventually you end up with BC≈7.0710678.

Now using the Law of Cosines, BD²=BC²+CD²-2(BC)(CD)cos C. Plugging in the numbers you end up with BD²≈50+100-2(7.0710678)(10)cos 45. Eventually you end up with BD=BC=7.0710678. Since triangle BCD is isosceles and∠C=45, ∠BDC=45 also. Now ∠ADB=180-45 which gives you the answer 135.

To make the calculation easier, assume AB = CD = x.

By using sin rule,

AB/(sinC) = AC/(sinB)

x/(sin45°) = AC/(sin105°)

AC = (x(√3 + 1))/2

AD = AC - CD

AD = (x(√3 - 1))/2

Again by using sin rule,

AB/sinC = BC/sinA

x/(sin45°) = BC/(sin30°)

BC = x/√2

By using Stewart Theorem,

BD² · AC = AB² · CD + BC² · AD - AD · CD · AC

BD² · ((x(√3 + 1))/2) = x² · x + (x²/2) · ((x(√3 - 1))/2) - ((x(√3 - 1))/2) · x · ((x(√3 + 1))/2)

BD² · ((x(√3 + 1))/2) = x³ + (x³(√3 - 1))/4 - x³/2

BD² · ((x(√3 + 1))/2) = (x³(√3 + 1))/4

BD² = x²/2

BD = x/√2

By using sin rule,

BD/(sinC) = BC/(sin(BDC))

(x/√2)/(sin45°) = (x/√2)/(sin(BDC))

Angle BDC = 45°

Angle ADB = 180° - (Angle BDC)

Angle ADB = 180° - 45°

Angle ADB = 135°

To find the solution of the question, we have to do the law o sines: 1) ∠ADB= α ,∠CBD= θ,so we conclude that ∠ADB= θ + 45º 2)Applying the law of sines on the triangle BDC: DC/sin θ = BD/sin45º so, sin θ = DC\cdot\sqrt{2} 3))Applying the law of sines on the triangle ABD:BD/sin 30º = AB/sin α so, sin α = AB/2\cdot\BD 4)solving the system, we found that θ=90º so ∠ADB= θ + 45º= 135º

Firstly let us take AB =1(x). Using the sine rule,a/sin(A)=b/sin(B). Therefore, we find that 1(x)/sin(45) * sin(30)=BC=0.7071(x). Now let us add a line from B to D: we know that CD=(x) and BC=0.7071(x). Using cos rule: a^2=b^2+c^2-2bc cos(A) to find DB^2 we input the values of CB,CD and <DCB into the formula to get 1(x)^2+0.7071(x)^2-2 1(x) 0.7071(x) cos(45)=0.5x^2. therefore, DB=squareroot(0.5)=0.7071. This is equal to BC, so the triangle CBD must be isosceles. Therefore the angle <CDB=<DCB=45. The angle <ADB=180-<CDB=135. Therefore the answer is 135.

Solution 1: Let $E$ be the foot of the perpendicular from $B$ to $AC$ . Since $\angle BCE = 45^\circ$ , so $BCE$ is an right isosceles triangle. Also, $ABE$ is a $30-60-90$ triangle. Thus $CE = BE = \frac{1}{2} BA$ . Since $CD = BA$ , it follows that $ED = \frac{1}{2} BA$ , or that $BED$ is also a right isosceles triangle. Thus $\angle BDE = 45^\circ$ and $\angle BDE = 180^\circ - \angle BDA = 135 ^\circ$ .

Solution 2: Let $\angle BDC = \alpha$ , then $\angle DBC = 135^\circ - \alpha$ .

Applying sine rule to triangle $ABC$ , we get $\frac { BC} {\sin 30^\circ } = \frac { AB} { \sin 45^\circ }$ .
Applying sine rule to triangle $BCD$ , we get $\frac {BC} { \sin \alpha } = \frac { DC } { \sin (135^\circ - \alpha ) }$ .

Since $AB = DC$ , hence $\frac{1}{ \sqrt{2}} = \frac{ \sin 30^\circ } { \sin 45^\circ} = \frac{BC}{AB} = \frac { BC} { DC} = \frac { \sin \alpha } { \sin(135^ \circ - \alpha ) } .$

Expanding this out, we get $\sqrt{2} \sin \alpha = \sin (135^\circ - \alpha) = \frac{\sqrt{2}} {2} \cos \alpha - ( - \frac { \sqrt{2} } {2} \sin \alpha)$ , or that $\tan \alpha = 1$ . Hence $\alpha = 45^\circ$ and so $\angle ADB = 135 ^ \circ$ .

Let AB = c and BC = a. By the sine rule, $\frac{c}{sin(45)} = \frac{a}{sin(30)}$ . In particular, we can rearrange this to $a^2 + a^2 = c^2$ .

Draw the line BD' with D' on AC such that BD' is perpendicular to BC. This gives an angle of BD'C = 45, so BD' = BC = a by isosceles-ness, so D'C = c by pythagoras, so D' and D are the same point.

From here, it is easy to obtain ADB = AD'B = 135 degrees.