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

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.

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 $$\angle ABD=48^\circ$$.

• thank you. may I know what software you use to make geometry diagrams? – rishikesh Jun 15 '20 at 14:25
• @rishikesh: It's CANVAS DRAW3 for Mac. – Christian Blatter Jun 15 '20 at 14:30
• @rishikesh There are a lot of software for that. Geometer's Sketchpad, Geogebra (Free), Desmos (Free/online). – Red Banana Jun 15 '20 at 19:19

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.

• 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. – rishikesh Jun 15 '20 at 14:52
• @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. – timon92 Jun 15 '20 at 16:31

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 $$B = 48^0$$

• I think you've made a mistake in the last step, the answer is 48 degrees. – rishikesh Jun 15 '20 at 18:42

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.)

• 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 – Dhanvi Sreenivasan Jun 15 '20 at 11:45
• Don't you use the cosine rule to get $\tan$? – A-Level Student Jun 15 '20 at 12:05
• No, just expand the $\sin(A-B)$ formula and re-arrange so that you get something of the form $p \sin A = q \cos A$ – Dhanvi Sreenivasan Jun 15 '20 at 12:08

Let $$\measuredangle ABC=x$$. Then $$\measuredangle BAD=114^\circ-x$$. By sine theorem we have: $$BD\stackrel{\triangle BAD}=AD\frac{\sin(114^\circ-x)}{\sin x} \stackrel{\triangle CAD}=AC\frac{\sin 48^\circ}{\sin 114^\circ}\frac{\sin(114^\circ-x)}{\sin x}\\ \implies \frac{\sin 48^\circ}{\sin 114^\circ}\frac{\sin(114^\circ-x)}{\sin x}=1 \implies \frac{\sin x}{\sin(114^\circ-x)}=\frac{\sin 48^\circ}{\sin 66^\circ} \implies x=48^\circ.$$