# Question about isosceles triangle

### Problem

$$\Delta ABC$$ is an isosceles triangle with $$AC=BC$$. $$BC$$ is extended to $$D$$ such that $$CD=AB$$. If $$\angle ADB=30^\circ$$, Find $$\angle ABD$$.

### My Progress

First of all, by using geogebra, I discovers that $$\angle ABD=20^\circ$$ for $$\angle ACB$$ being obtuse and $$\angle ABD=60^\circ$$ for $$\angle ACB$$ being acute.

I also try by extending $$AC$$ to $$E$$ such that $$CE=AB$$. Then $$ABED$$ would be an isosceles trapezium. I am not sure whether it helps.

Note: In what follows, we prove that $$\angle ABD=20^{\circ}$$ when $$\angle ACB$$ is obtuse.

(The case $$\angle ABD=60^{\circ}$$ when $$\angle ACB$$ is acute can be proved in exactly the same way.)

Draw equilateral triangle $$OAC$$ as shown.

Join OD.

Let $$\angle ABD = \theta$$.

Note that

$$(1)$$ If we use $$O$$ as the center, $$OC$$ as the radius and draw a circle, then $$\angle AOC= 2 \times \angle ADC \implies D$$ lies on the circle.

$$(2)$$ $$\therefore OD=OC=CA=CB$$ and $$CD=AB.$$

$$(3)$$ $$\Delta OCD \cong \Delta CBA$$. (SSS)

$$(4)$$ Hence $$\angle OCD=\angle ABC =\theta.$$

$$(5)$$ $$\therefore \angle ACO=3\theta.$$

$$(6)$$ $$\therefore 3\theta = 60^{\circ}$$ and $$\theta = 20^{\circ}.$$

If allowed to use trigonometry, assuming that $$\angle ABC=x$$, then:

$$\frac{\sin \angle DAC}{\sin 30^{\circ}}=\frac{\sin (30^{\circ}+2x)}{\frac{1}{2}}=\frac{DC}{AC}=\frac{AB}{AC}=\frac{\sin 2x}{\sin x} \\ \implies \sin (30^{\circ}+2x)=\cos x \\ \implies 30^{\circ}+3x=90^{\circ} \ or \ 30^{\circ}+x=90^{\circ} \implies x=20^{\circ} \ or \ x=60^{\circ}.$$

• the figure does not show this. Are you sure about your solution? Jul 17, 2023 at 10:40
• @sirous $60^{\circ}$ is very easy to verify, and according to the figure posted by the OP, $20^{\circ}$ makes sense ... Jul 17, 2023 at 10:47
• @sirous As mentioned by the OP, the two answers come from the angle $ACB$ being acute or obtuse. Jul 17, 2023 at 10:49
• +1,Thanks for clarification. I forgot that triangle may be acute. Jul 17, 2023 at 12:21

Hint: You can also use this figure , As shown ,reflect AC and DB about AD and connect C' to B nd B to B'. In triangle BDC', A is where bisectors of angles C', B and D meet and triangle DBB' is equilateral , so $$\angle B'BD=60^o$$. We have:

$$\angle DBA=\angle ABC'$$

Quadrilateral $$C' A'_2 B A$$ is a kite symmetric about CB , therefore:

$$\angle C'BB'=\angle DBA=\angle ABC'=\frac {60}3=20^o$$

You can apply the same method to acute triangle.