A Strange Star in a Strange Shape

A $7$ - pointed star is called a heptagram (or a septagram). There are two types of heptagrams possible $\{7/ 2\}$ and $\{7/ 3\}.$ For more info, click here

The above drawn heptagram is a $\{7/2\}$ heptagram.

For any $\{7/2\}$ heptagram , the sum of the angles of its vertices is constant. I shall now prove it to be $540^\circ$ .

I have labeled the heptagram as following. Notice that $IJKLMNP$ is a heptagon. Heptagon

Let the required sum $\angle A + \angle B + \angle C+ \angle D + \angle F + \angle G+ \angle H$ be $x.$

We can make seven quadrilaterals by choosing 3 non-adjacent vertices, of the heptagram and one corresponding side of the heptagon.

One such quadrilateral, $ANFC$ is highlighted in the diagram. As the sum of interior angles of quadrilateral is $360^{\circ}$ , We get the following equations for each quadrilateral.

$\angle A + \angle G + \angle C + \angle GLC = 360^{\circ}$

$\angle A + \angle C + \angle F + \angle ANF = 360^{\circ}$

$\angle A + \angle D + \angle G + \angle AJD = 360^{\circ}$

$\angle B + \angle H + \angle F + \angle FKB = 360^{\circ}$

$\angle B + \angle G + \angle D + \angle BPG = 360^{\circ}$

$\angle B + \angle H + \angle D + \angle LMN = 360^{\circ}$

$\angle C + \angle H + \angle F + \angle JIP = 360^{\circ}$

By adding these equations, we get

$3x + \angle GLC + \angle ANF + \angle AJD + \angle FKB + \angle BPG + \angle LMN + \angle JIP = 2520^{\circ}$

Notice $\angle GLC + \angle ANF + \angle AJD + \angle FKB + \angle BPG + \angle LMN + \angle JIP$ is sum of interior angles of heptagon, ie $900^{\circ}$ .

Solving for $x$ , we get,

$3x + 900^{\circ}= 2520^{\circ}$

$3x = 1620^{\circ}$

$\boxed{x = 540^{\circ}}$

We have 7 yellow triangles, the base angles of those 7 yellow triangles form the exterior angles of heptagon.The sum of exterior angles are 2 (360) = 720 So <A + <B + < C+<D +< F+<G +< H = 7(180) – 720 = 540

Start traversing the star from A in the counter-clockwise direction -

Let $\angle OAG = \theta$ .

Then $\angle AGO = \theta'$ it's compliment.

Since GC bisects $\angle AGD$ , $\angle OGD = \theta'$ as well.

$\angle GDB = \theta$ compliment of compliment

$\angle DBH = 90°$ and $\angle BHF = 90°$ are given.

Had we started traversing from A in the clockwise direction, we would have got -

$\angle OAC = \theta$

$\angle ACO = \theta'$

$\angle OCF = \theta'$

$\angle CFH = \theta$

Adding them all we have $\theta + \theta' + \theta' + \theta + 90° + 90° + \theta + \theta' + \theta' + \theta = 4\theta + 4\theta' + 180° = 540°$

i just got it without solving... haha ... crap that was just a big mistake

In triangle AGC, A+(G/2)+(C/2)=180. Now, F+(C/2)=90 & D+(G/2)=90. Hence, A+B+C+D+E+F+G+H=540.