Ok....so I have made this post a bit better (I think so )

Define:
reflections of circumcenter of , over
foot of perpendiculars from to
midpoints of
midpoints of
Nine-point center of

Note: We'll also include some basic properties of nine-point circle too!

Pre-Eliminaries

Lemma 1: Reflection of orthocenter over any side lies on the circumcircle

Proof:

Lemma 2:Reflection of orthocenter over any midpoint of a side lies on the circumcircle

Proof:

Lemma 3: Perpendiculars from vertex to concur at the circumcenter of

Proof: Note that are the excenters WRT , hence, they concur at the bevan point WRT which is indeed the circumcenter WRT

Lemma 4: are infact the circumcenters of

Proof:

Lemma 5: The Nine-Point Centers of and are the same

Proof:


Main(Basic) Properties

Property 1:Circles are congruent

Proof:

Property 2: is the orthocenter WRT and

Proof:

Property 3: is the circumcenter WRT and

Proof:

Property 4: (Nine-Point Circle WRT ) bisects any segment from the orthocenter to any point on the circumcircle of

Proof:

Property 5: Nine-point center is the circumcenter WRT and
(Restatement) The perpendicular bisectors of and concur at

Proof:

Property 6: Quadrilaterals and are parallelograms

Proof:

Property 7: and are congruent triangles

Proof:

Property 8: If and , then,

Proof: See this, (Post #13, Lemma 1)

Property 9: and are concurrent at and bisects

Proof:

Property 10: and are Rhombus

Proof:

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Let's See these basic lemmas and properties in action!

Q1)(Source= BAMO 2013/P5) Solution:
Q2)(Source=HKMO Round 1 2018/P6) Solution:

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And Lastly, here are some problems to try!

Problems

1# Macedonia National Olympiad 2015/P1 Let and be altitudes in . Let be the perpendicular lines from vertices to respectively. Prove that are concurrent lines.

2# APMO 2004/P2 Let be the circumcenter and the orthocenter of an acute triangle . Prove that the area of one of the triangles , and is equal to the sum of the areas of the other two.

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(Let me know other problems that can be solved using the lemmas and properties stated in this post, below!)