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
3 years, 10 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 $XY || BC$.

- 3 years, 10 months ago

@Alan Yan How u so genius?

- 3 years, 10 months 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.

- 3 years, 10 months ago

Hints: 1. Property of circumcenter 2. Area

- 3 years, 10 months ago

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

- 3 years, 10 months ago

Exactly :P

- 3 years, 10 months 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

- 3 years, 10 months 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

- 3 years, 10 months 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 $\angle BSD = \angle BAC$.

- 3 years, 10 months ago