Find $\angle CAD$ if $\triangle ABC$ is right angled at $B$, $\angle BAD = 30^\circ, \angle ADB = \angle ADC = 15^\circ$

Question

Find angle $\theta$ in the below diagram.

This is a question that was brought to me by a high school student.

While I came up with a trigonometric solution and a synthetic solution, I am posting here to see more solutions that others come up with (esp. other synthetic solutions)

My immediate solution involved combination of a simple construction and trigonometry, basically knowing that $\tan 30^\circ = \dfrac{1}{\sqrt3}$ and $\tan 15^\circ = 2 - \sqrt3$.

We draw perp from $D$ to $AB$ extend and $BC$ extend. We also note that $\angle DBC = \angle DBE = 45^\circ$. If $AB = x, BE = y$, we find $x$ in terms of $y$. We next find $CF$ in terms of $y$ and subtracting from $y$ gives us $BC$ and we show $BC = x$.

Then in search of a synthetic solution, I drew a few more lines and as $FE$ is perpendicular bisector of $BD$, $BG = GD, BH = HD$.

So we see that $AB = BG$ and by A-A-S, $\triangle BHC \cong \triangle BHG$ which leads to $AB = BC$ and we have $\theta = 15^\circ$.

Look forward to more interesting solutions.

Answer 1

Reflect $B$ across $AD$ to $F$. Then $\angle FDA = \angle CDA = 30^{\circ}$ so $F,C,D$ are colinear.

• Since $\angle CBF = 30^{\circ}$ and $\angle BFC = 75^{\circ}$ we see that $BF = BC=:r$.
• Since $AB = AF$ and $\angle BAF = 60 $ we have $AB = AF = BF =r$.
• So $A,F,C$ are on a circle with center at $B$ and radius $r$, so $$\angle FAC = {1\over 2}\angle CBF = 15^{\circ}$$ and so $ \angle CAD = 15^{\circ}$

Answer 2

Draw the bisector of $\hat A$, it touches BD at E.Triangle AED is isosceles. Also $\angle DEA=\angle ADC$ that means CD||AE .$\angle BCE=\angle BAD=30^o$, also $CB \bot AB$, therefore $CE\bot AD$ that is AEDC is a rhombus and triangle ACD is isosceles therefore $\theta=15^o$

Answer 3

Rotate $D$ around $C$ for $-60^{\circ}$ in to $F$ and let $E$ be intersection point of $AD$ and $BC$. Then $CDEF$ is cyclic and

• $ABE$ is congruent to $FBE$ (a.s.a)
• $CDB$ is congruent to $FDB$ (s.a.s) so $\angle CBF = 90^{\circ}$

and so $AEC$ congruent to $FEC$ (s.a.s). Thus $\angle EAC = \angle EFC = 15^{\circ}$.

Answer 4

Sine rule for $\Delta ABD$: $$\frac{AB}{\sin 15^\circ}=\frac{BD}{\sin 30^\circ} \Rightarrow BD=\frac{AB}{2\sin 15^\circ}$$ Sine rule for $\Delta BCD$: $$\frac{BC}{\sin 30^\circ}=\frac{BD}{\sin 105^\circ} \Rightarrow BC=\frac{BD}{2\sin 105^\circ}=\frac{BD}{2\cos 15^\circ}=\frac{AB}{4\sin 15^\circ\cos 15^\circ}=AB$$ Hence, the triangle $ABC$ is right angled and isosceles, implying $\theta=15^\circ$.

Answer 5

Here's my solution to this, sorry for being too late.

I'll add some explanation here.

1. line BK meets AD such that AB=BK=KD.
2. Map point B onto point B' and rotate Triangle DCB such that the new triangle DCB' formed it congruent to triangle DCB, and <BDB'=60.
3. Join point B' and B, the resulting Triangle BDB' is an equilateral triangle.
4. Notice that Triangle BCB' is congruent to Triangle BKD via the ASA property. Therefore, line segment B'C=BC=BK=AB=KD
5. Above implies that Triangle CBA is an isosceles right triangle, therefore the unknown angle is 45-30=15.