1) if S is the circumcentre of a \(\triangle ABC\) Then prove that \(\angle BSD = \angle BAC\) given that D is the mid point of BC.

2) Let P be any point inside a regular polygon. If \(d_i\) is the distance of P from the \(i^{th}\) side, prove that \(d_1+d_2+d_3+....+d_n\) is constant.

3) I is the incentre of Triangle ABC.X and Y are the feet of the perpendiculars from A to BI and CI. Prove XY || BC

I liked these problems, so i thought that I should share them with the community! :D

Edit: Xuming Liang is forbidden to comment.

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## Comments

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TopNewestQuestion 3: Denote the point where \(BI\) hits \(AC\) as \(E\). Denote the angle measures as \(A, B, C\) according to the vertices of the triangle. All angles in this solution are in degrees.

By Exterior Angle Theorem, we have that \(\angle BEA = C + \frac{B}{2} \). Since \(\angle AYE = 90\), we have that \(\angle YAE = 90 - C - \frac{B}{2} \).

Connecting \(A\) to \(I\), we know that \(AI\) is the angle bisector of \(\angle A\), which implies that \(\angle IAE = 90 - \frac{B}{2} - \frac{C}{2} \). This implies that \(\angle IAY = \frac{C}{2} \).

Now notice since \(\angle AXI = \angle AYI = 90\), we have that quadrilateral \(AXIY\) is cyclic. This means that if we connect \(X\) and \(Y\), we have that \(\angle IAY = \angle IXY = \frac{C}{2} \). However we also know that \(\angle XCB = \frac{C}{2} \). Therefore by alternate interior angles, we know that \(XY || BC\).

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@Alan Yan How u so genius?

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Question 2. Partition the polygon into triangles. Let \(m\) be the side length and let \(A\) be the total area. The area of the triangles will be \( \frac{1}{2}md_1 , \frac{1}{2}md_2 , \frac{1}{2}md_3 , ... \). Adding these areas yield the total area.

This implies that \[\frac{1}{2}m(d_1+d_2+...+d_n) = A \implies \sum{d_i} = \frac{2A}{m} \] which means the sum of the distances is constant.

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Hints: 1. Property of circumcenter 2. Area

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You are not allowed to comment on such posts :) :P

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Exactly :P

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@Xuming Liang Geom god! _/_

The first one is quite simple. The second one is interesting :)

But you man! These probs are a left hand's play for you, provided you are right handed. :3

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in triangle BDC and CDS,

BS = SC

BD = DC

and angle d is 90

so both triangles are congruent and then angle BSD = angle CSD

also 2.angleBSD = 2.angleBAC

hence proved

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Question 1. \(\angle BSD = \frac{1}{2}\angle BSC \) (Perpendicular Bisector.)

\(\angle BAC = \frac{1}{2}\angle BSC \) (Central angle and inscribed angle substending the same arc.)

Therefore \( \angle BSD = \angle BAC\).

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