ABCD is a square. P is an arbitrary point on DC. <PAQ = 45°, Q is on BC. Join P & Q. & Prove that <AQB= <AQP, also <APD = < APQ?

Answer 1

The diagonals of the square bisect it's angles and each other.

M is the intersection of PQ and diagonal AC

THEN angles DAP = CAQ = t and angles PAM and QAB = 45-t, since the corner angles are 45 each.

Let angle AQB = x, AQM = z, AMQ = lambda, AMP = alpha, APQ = w, and APD = y.

We wish to prove that x=z and y=w, and such shall be done by contradiction, by supposing that x and z are distinct angles, as are y and w.

Per triangle AQM, t + z + lambda = 180

Per triangle AMP w + alpha - t = 135

then w + z + alpha + lambda = 315.

However, w + z + alpha + lambda = 135 per triangle APQ .

This contradiction proves that the original assumption is false.

Specifically , it is NOT true that neither pair of angles are unequal.

Therefore, one pair of angles must be equal.

Without loss of generality (WLOG), Repeating this argument for angles ADP = APM = y, also produces

the same contradiction which forces the other pair of angles to be equal.

(You get a system of 5 unknown and 5 equations which produces the contradiction, upon solution, that

2w + lamba = 135 = 45).

Therefore, BOTH pairs of angles as stated must be equal to each other.

As before, ADP = APM = y, with angles AQB = AQM = x.

There are now 5 unknowns: t,x,w,lamba, and y

y + t = 90 from triangle ADP

y + lambda - t = 225

t + w +x = 180 per triangle AMQ

x-t = 45 per triangle ABQ

Per quadrilateral AQPD, x + t + 2y = 225

solves the first equation for t , gives t = 90-y.

BUT, by quadrilateral ABQP

45 + t + 90 +2x + y = 360

135 + 2x + t+ y = 360

2x + t + y = 225

2x + (90- y) + y = 225

2x + 90 = 225

2x = 135

x = 67.5

which forces t=22.5

which forces w = 90

which forces lambda = 90

which forces y = 67.5

Therefore x=y = 67.5

which concludes the proof

[end of proof]