KOMATH

<Qua-2025-770>, HINT

Let a,b,c,d be the x-coordinates of point A,B,C,D and

A',B',C',D' be the x-coordinates of points A',B',C',D'

(A'+a)/2=b ---> A'=2b-a, B'=2c-b, C'=2d-c, D'=2a-d, Add up all

A'+B'+C'+D'=a+b+c+d. Suppose above origin (0,0)

C'+2D'=4a-c A'+2B'=4c-a by arranging 15a=A'+2B'+4C'+8D'

Similarly b,c,d are defined with given A',B',C',D'.

<Qua-2025-770>, general, construction

<Tri-2025-2223>, hint

AH cuts BC in H'.

Since AD is the bisector of <A=90 , APDQ is a square and BD/CD=AB/AC

Triangles ABC,PBD, QDC are similar.

By Ceva's theorem (AP/PB)x(BH'/H'C)x(CQ/QA)=1

You can find AH' is altitude.

<Tri-2025-2223>

<Tri-2025-2222>, hint

Let area of triangle ABC be S

Since AB=A'B (ACA')=2S. Similarly (CA'C')=4S (CB'C')=2S, (CA'B')= S

Hence (A'B'C')=7S , B" on A'C', B'B" be altitude of A'B'C'. A" on B'C', A'A" be altitude

Let B"B*= 4/7 B"B', through B* parallel to B'C' draw line L'

Since (CA'C')=4S, L pass through point C

Let A"A*=2/7 A"A', through A* parallel to B'C' draw line L'. Since (C'B'C)=2S

L' pass through C too.

Finally Point C is the intersection of line L and L'.