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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
2 years, 1 month 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 $$XY || BC$$.

- 2 years, 1 month ago

@Alan Yan How u so genius?

- 2 years, 1 month ago

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 $$\angle BSD = \angle BAC$$.

- 2 years, 1 month ago

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.

- 2 years, 1 month ago

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

- 2 years, 1 month ago

Hints: 1. Property of circumcenter 2. Area

- 2 years, 1 month ago

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

- 2 years, 1 month ago

Exactly :P

- 2 years, 1 month ago

@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

- 2 years, 1 month ago