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In triangle $ABC$, $\angle C = 48^\circ$. $D$ is any point on $BC$, such that $\angle CAD = 18^\circ$ and $AC = BD$. Find $\angle ABD.$

I tried to make some constructions: draw a line through $D$ parallel to $AC$; draw a line through $C$ which makes $66$ degrees with $AC$. None of them have been useful. It feels like I am trying to force the formation of congruent triangles, and that doesn't work. Please help.

Diagram created using GeoGebra

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Draw in black the triangle $ABC$ with angle $48^\circ$ at $C$, and $AD$ with angle $18^\circ$ at $A$. This ensures the blue angles $114^\circ$ and $66^\circ$. Choose $D'\in BC$ such that $|D'C|=|BD|=|AC|$. This gives the red angles $66^\circ$ and $66^\circ-18^\circ=48^\circ$. This implies that $|AD'|=|AD|$. The triangles $AD'C$ and $ADB$ are therefore SAS-congruent with the same angles at $D'$ and $D$. This shows that the green angle $\angle ABD=48^\circ$.

enter image description here

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  • $\begingroup$ thank you. may I know what software you use to make geometry diagrams? $\endgroup$ – rishikesh Jun 15 '20 at 14:25
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    $\begingroup$ @rishikesh: It's CANVAS DRAW3 for Mac. $\endgroup$ – Christian Blatter Jun 15 '20 at 14:30
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    $\begingroup$ @rishikesh There are a lot of software for that. Geometer's Sketchpad, Geogebra (Free), Desmos (Free/online). $\endgroup$ – Red Banana Jun 15 '20 at 19:19
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Let $E$ be a point such $ED=AD$ and $EB=CD$ and that $A, E$ lie on the same side of line $BC$. Then $\triangle DEB \equiv \triangle ADC$ so $\angle BDE = \angle CAD$ and therefore $$\angle EDA = \angle BDA - \angle BDE = \angle DCA + \angle CAD - \angle BDE = \angle DCA = 48^\circ.$$ Since $ED=AD$, we have $$\angle AED = 90^\circ - \frac 12 \angle EDA = 90^\circ - \frac 12 \cdot 48^\circ = 66^\circ.$$ But also $\angle DEB = \angle ADC = 180^\circ - 48^\circ - 18^\circ = 114^\circ$. Hence $\angle AED + \angle DEB = 180^\circ$ and therefore $E$ lies on $AB$. It follows that $$\angle ABD = \angle EBD = \angle DCA = 48^\circ.$$

Special thanks to Calum Gilhooley for providing a picture.

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  • $\begingroup$ how do we know that point E exists? if we draw circles centred at A and B with the chosen radii, it is not immediately apparent to me that they intersect. $\endgroup$ – rishikesh Jun 15 '20 at 14:52
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    $\begingroup$ @rishikesh These circles intersect because the sum of their radii is bigger than the distance between the centers. This follows from triangle inequality in $ACD$ and from the assumption that $AC=BD$. The intended definition of $E$ is: build a triangle $DEB$ congruent to $ADC$. It is clear that this exists. $\endgroup$ – timon92 Jun 15 '20 at 16:31
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One way to solve it would be using sine rule. In ACD, we have

$$\frac{\sin 114}{AC} = \frac{\sin 48}{AD}$$

In triangle ABD, we have

$$\frac{\sin(114-B)}{BD} = \frac{\sin B}{AD}$$

If you solve these equations, using AC=BD, we have $$\tan B = \frac{\sin 48}{1+\cot 114\cdot\sin48}$$

$$\implies \tan B = 1.1107$$

Hence $B = 48^0$

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  • $\begingroup$ I think you've made a mistake in the last step, the answer is 48 degrees. $\endgroup$ – rishikesh Jun 15 '20 at 18:42
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Ok, here's my method, it's VERY confusing and messy, so there's probably a better method. Using the sine rule: $$\frac{AC}{\sin114}=\frac{AD}{\sin 48}$$ so $$AC=BD=\frac{AD\sin114}{\sin48}$$ Cosine rule: $$AB^2=AD^2+\frac{AD^2\sin^2 114}{\sin^2 48}-2\frac{AD^2\sin 114\cos 66}{\sin 48}=AD^2 (1+\frac{\sin^2 114}{\sin^2 48}-2\frac{\sin 114\cos 66}{\sin 48})$$ So $$AB=AD \sqrt{1+\frac{\sin^2 114}{\sin^2 48}-2\frac{\sin 114\cos 66}{\sin 48}}$$ Sine rule (again): $$\frac{AB}{\sin66}=\frac{AD}{\sin ABD}$$ so $$\frac{AD \sqrt{1+\frac{\sin^2 114}{\sin^2 48}-2\frac{\sin 114\cos 66}{\sin 48}}}{\sin66}=\frac{AD}{\sin ABD}$$ Cancelling $AD$ from both ides and rearranging to find $ABC$ should get you your answer. I hope that helped! (It'd probably help to draw a sketch.)

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    $\begingroup$ Rather than using cosine rule, I think sine rule with some re-arrangement would make a better approach, just check my solution if it's on the same lines $\endgroup$ – Dhanvi Sreenivasan Jun 15 '20 at 11:45
  • $\begingroup$ Don't you use the cosine rule to get $\tan$? $\endgroup$ – A-Level Student Jun 15 '20 at 12:05
  • $\begingroup$ No, just expand the $\sin(A-B)$ formula and re-arrange so that you get something of the form $p \sin A = q \cos A$ $\endgroup$ – Dhanvi Sreenivasan Jun 15 '20 at 12:08
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Let $\measuredangle ABC=x$. Then $\measuredangle BAD=114^\circ-x$. By sine theorem we have: $$ BD\stackrel{\triangle BAD}=AD\frac{\sin(114^\circ-x)}{\sin x} \stackrel{\triangle CAD}=AC\frac{\sin 48^\circ}{\sin 114^\circ}\frac{\sin(114^\circ-x)}{\sin x}\\ \implies \frac{\sin 48^\circ}{\sin 114^\circ}\frac{\sin(114^\circ-x)}{\sin x}=1 \implies \frac{\sin x}{\sin(114^\circ-x)}=\frac{\sin 48^\circ}{\sin 66^\circ} \implies x=48^\circ. $$

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