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.

Easy Math Editor

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Comments

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 - 5 years, 9 months ago

@Alan Yan How u so genius?

Mehul Arora - 5 years, 9 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.

Alan Yan - 5 years, 9 months ago

Hints: 1. Property of circumcenter 2. Area

Xuming Liang - 5 years, 9 months ago

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

Nihar Mahajan - 5 years, 9 months ago

Exactly :P

Mehul Arora - 5 years, 9 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

Mehul Arora - 5 years, 9 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

Dev Sharma - 5 years, 9 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 $\angle BSD = \angle BAC$ .

Alan Yan - 5 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$