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For reference: In triangle $ABC$, let $D$ and $E$ be the intersections of the bisectors of angles $ABC$ and $ACB$ with sides $AC$ and $AB$ respectively. Knowing that the measures, in degrees, of angles $BDE$ and $CED$ are equal to $24$ and $18$ respectively, the difference, in degrees, between the measures of the two smallest angles of this triangle is equal to?

My progress: My drawing

enter image description here

$\angle EID = 180-18-24 = 138^o=\angle CIB \implies \angle DIC = 42^o=\angle BIE\\ \triangle CIB\alpha+\theta+138=180 \therefore \alpha +\theta = 42^o\\ \triangle ABC: \angle A +2\alpha +2\theta =180^o \implies \angle A= 180 - 2(\alpha+\beta)=180 -2(42) \therefore \angle A = 96^o $

...???

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2 Answers 2

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I believe, there is some trick to simplify solution of the problem with given angles. But I haven't found it.

My solution can be used for any values of given angles and is based on counting projections of edge FG on direction of edge ED and direction perpendicular to edge ED.

picture to the solution

The point F is point symmetric to E about BD and the point G is point symmetric to D about CE. Then EG = ED = DF, angle FDE is 48°, angle GED is 36°.

To solve the problem we need to find $\alpha$ which is $18°+x$ in my picture. So we need to find $x$. The $\tan x$ can be found as shown in the picture.

For given angles result $x=18°$ requires hard calculations. I was doing these by founding $\sin 18°=\frac{\sqrt{5}-1}{4}$ and using this expression in next calculations.

The final answer is $2\alpha-2\theta=4x-12°=60°$

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  • $\begingroup$ Other way. From sine rule for triangles FIG and EID follows: $\sin(42°+\alpha) \sin 18° = \sin(84°-\alpha) \sin 24°$. Expanding and simplifying gives $\tan \alpha = \frac{1+\cos 72°-\cos 24°}{\sin 72°-\sin 24°}$ $\endgroup$ Mar 2, 2022 at 9:27
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enter image description here

Hints: In figure we have:

$\overset{\large\frown}{EG}=24\times2=48^o$

G is midpoint of arc EF so:

$\overset{\large\frown}{EF}=48\times2=96^o$

$\overset{\large\frown}{EF}=\overset{\large\frown}{EA}=\overset{\large\frown}{AK}=2\times 96=192^o$

where K is between D and C on the big circle.

You found $\angle A=96^o$

Using these data you have to show:

$\angle A=\angle E=\angle F=96^o$

Which gives:

$$\angle FCA=360-3\times 96=72$$

so:

$\angle B=(180-96=84)-72= 12^o$

So :

$$\alpha-\theta=36-6=30^o$$

The key point is that CD is perpendicular on EF and $\angle A=\angle E=\angle F=96^o$ and FC is bisector of $\angle E$ which means $\angle AEC=48^o$. In this way:

$\angle AED=30^o|rightarrow \angle ACE=30^o$

which finally gives $\hat{C}=72^o$

Note that EF and EA are tangent on inscribed circle and quadrilateral ACEF is symmetric about EC, this is the reason for $\angle F=\angle A=96^o$.Also BI is perpendicular on EF because the rays of angle $\hat B$ are tangent on inscribed circle and also BI is bisector of that angle.Point F is mirror of point A about CE.

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  • $\begingroup$ Thanks, but can you explain better? How did you draw the larger circle? Why is G the midpoint of EF? I don't understand the point K... $\endgroup$ Feb 28, 2022 at 19:26
  • $\begingroup$ understood it like this... Completing the circumscribed quadrilateral we will have the point $T$ of tangency that will be perpendicular to the center of the circumference. Therefore the $\angle LEI = 48^o \implies \angle AED = 30^o~ and~\angle ACE = 36^o \implies \angle C = 72^o$ Is correct? $\endgroup$ Feb 28, 2022 at 23:44
  • $\begingroup$ @sirous: How big circle was drawed: as circle with center I and radius IE or as circle with center I and radius IA? $\endgroup$ Mar 1, 2022 at 12:13
  • $\begingroup$ @sirous: What is point F: such point of BC that EF is tangent to inscribed circle, or such point of BC that BE=BF ? $\endgroup$ Mar 1, 2022 at 12:29
  • $\begingroup$ @petaarantes: (about your comment) you still need to prove that tangent EL is perpendicular to BI to complete solution. $\endgroup$ Mar 1, 2022 at 12:46

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