Question

Can anyone help me prove the theorem 49 by using the following hint( Circle)?

Strong Right Angle Theorem 49. All right angles are congruer Consequently, a right angle is congruent to another angle if and only if the other angle is also a right angle.) Since you have already proven Theorem 39, it remains only to prove the converse: i f LX and ZX' are right angles, then they are congru- ent. To do this, choose points P and Y with X between them and Y

Here is Theorem 39, can't use like right angles are 90 degrees, or that triangles are 180 degrees.

Answer 1

According to theorem 39 (weak right angel theorem) a angle that is congruent to aright angle is also right angle given: - In circle c, the construction of circle c is copies y & Y^1 are centers P_y & P^1y^1 are radius of circle & c^1 respectively to prove: delta xyz almostequalto delta x^1 y^1 z^1 proof zy = z^1y^1 (radius of circle) yQ = y^1 Q^1 (radius) px = p^1x^1 (distance from the x&y^1 to the end of the arc would be equal) therefore xy = x^1y^1 (xy = py - px (py is radius) \r\n (x^1 y^1 = p^1 y^1 - p^1 x^1(p^1 y^1 is radius) therefore angle xyz = angle x^1 y^1 z^1 (angle subtended by the equal are F_2 & p_2 to the centers is also equal) therefore delta xyz almostequalto delta x^1 y^1 z^1 by SAS criteria [xy = x^1 y^1 angle xyz = anglex^1 y^1 z^1 angle zy = angle z^1 y^1] \r\n