Draw square PQRS with side length a. Choose a point on SR other than the endpoints and label it T. (See attachment for diagram.)

We are given that U is a point on QR such that `/_STP ~= /_PTU` . Reflect line ST over line PT to get line TS';...

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Draw square PQRS with side length a. Choose a point on SR other than the endpoints and label it T. (See attachment for diagram.)

We are given that U is a point on QR such that `/_STP ~= /_PTU` . Reflect line ST over line PT to get line TS'; line TS' intersects RQ at U. Note that reflections preserve angles so that PT is the angle bisector of `/_STU` .

Now reflect PS over PT to get PS'. Note that S' lies on TU, the length of PS' is a, and `/_PS'T=90^@` , since reflections preserve angles and lengths.

Let `m/_STP=m/_S'TP=x` . Note that `Delta PQU ~= Delta PS'U` by HL. Let `m/_PUQ=m/_PUS'=y` , since corresponding angles are congruent.

Now by angle addition, we have `m/_RTU=180-2x` .

We have two expressions for the measure of angle RUT:

`m/_RUT=90-(180-2x)=2x-90` , since the angles of a triangle sum to 180. But `m/_RUT=180-2y` . So 180-2y=2x-90 or x+y=135.

Now consider `Delta PTU` . We see that `m/_TPU+x+y=180^@ ==> m/_TPU=180-135=45^@` .