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Consider two right angle triangles ABC & PQM with hypotenuse AC and PM and 1 side BC & QM equal.
BC = a and QM = p, a = p. AC = h = PM.
Move ∇PQM so that Q lies above B, and rotate triangle so QM lies above the line BC then M must lie above C as BC = QM.
Extend line AB above A. Draw an arc of radius h centred on C and it will intersect this extended line at A since AC = h = QM.
∴ ∇ABC and ∇PQM are congruent (equal).
∴ ∇ABC and ∇PQM are congruent (equal).
Mirror Triangle
Mirror Triangle
Above only one possible solution was covered, where the radius h intersected the line AB extended above the line BC.
If the line BC is extended further below the line BC and the circle continued it will intersect again at position R. Join BR and CR and there is another right angle triangle with side equal to a and hypotenuse equal to h.
Call this ∇ RST, and it is the mirror image of ∇ PQM.
This can be treated in same manner as Triangles with 2 sides and inc. angle equal, but requires adopting a 3D space in what is supposed to be a 2D environment.
Rotate ∇ RST about axis ST so that it lies above ∇ ABC.
As before draw an arc of radius h centred on C and it will intersect this extended line at A since AC = h = RT.
∴ ∇ ABC and ∇ PQM are congruent (equal).
∴ ∇ ABC and ∇ PQM are congruent (equal).
Alternative Proof for Mirror Triangle
Alternative Proof for Mirror Triangle
Consider ∇s ABC & RST with ST overlaying BC.
If S is over B then T is over C as BC = CT = a
AC = RT = h
∠ ABC = ∠ RST = 1 rt. ang.
∴ AR is a straight line
∠ CAB = ∠ TRS (∇ ACR is isos. AC = RT)
∴ ∠ ACB = ∠ RCT (sum tri. ang.s = 180º)
So ∇s ABC & RST are equal (2 sides and incl angle equal)
So ∇s ABC & RST are equal (2 sides and incl angle equal)
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