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Good proof problems

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.

Note by Mehul Arora
1 year, 6 months ago

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Question 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 \(XY || BC\). Alan Yan · 1 year, 6 months ago

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@Alan Yan @Alan Yan How u so genius? Mehul Arora · 1 year, 6 months ago

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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 \( \angle BSD = \angle BAC\). Alan Yan · 1 year, 6 months ago

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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. Alan Yan · 1 year, 6 months ago

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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 Dev Sharma · 1 year, 6 months ago

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Hints: 1. Property of circumcenter 2. Area Xuming Liang · 1 year, 6 months ago

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@Xuming Liang You are not allowed to comment on such posts :) :P Nihar Mahajan · 1 year, 6 months ago

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@Nihar Mahajan Exactly :P Mehul Arora · 1 year, 6 months ago

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@Xuming Liang @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 Mehul Arora · 1 year, 6 months ago

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