In fig. the bisectors of the angles $\angle DAC$, $ \angle EBD$, $\angle ACE$, $\angle BDA$ and $\angle CEB$ intersect at one point. Prove that the bisectors of the angles $\angle TPQ$, $\angle PQR$, $\angle QRS$, $\angle RST$ and $\angle STP$ also intersect at one point.
Problem
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Tags: geometry, angle bisector, concurrency, concurrent
Spacetime_of_Pi
03.05.2021 21:52
I have a good idea for this problem.
The equation for finding the sum of the interior angles is ( n - 2 ) * 180 when n = # of sides.
further hintThe sum of all the interior angles is 540 and the sum of all int angles in a 4 sided is 360. For triangles it is 180 and so on.
further further hintTry creating a system of equations from all the sums of the angles you now know.
This could be wrong, can someone check?
sunken rock
04.05.2021 12:40
If the angle bisectors of $\angle A, \angle B,\angle C,\angle D,\angle E$ concur at $I$, then which is the incenter of $\triangle ADQ$? Best regards, sunken rock