6
$\begingroup$

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.

$\endgroup$

3 Answers 3

7
$\begingroup$

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.)

enter image description here

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 $\theta = 20^{\circ}.$

$\endgroup$
7
$\begingroup$

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

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

$\endgroup$
4
  • $\begingroup$ the figure does not show this. Are you sure about your solution? $\endgroup$
    – sirous
    Jul 17, 2023 at 10:40
  • $\begingroup$ @sirous $60^{\circ}$ is very easy to verify, and according to the figure posted by the OP, $20^{\circ}$ makes sense ... $\endgroup$ Jul 17, 2023 at 10:47
  • $\begingroup$ @sirous As mentioned by the OP, the two answers come from the angle $ACB$ being acute or obtuse. $\endgroup$ Jul 17, 2023 at 10:49
  • 1
    $\begingroup$ +1,Thanks for clarification. I forgot that triangle may be acute. $\endgroup$
    – sirous
    Jul 17, 2023 at 12:21
3
$\begingroup$

enter image description here

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .