Question

in triangle PQR, S is a point on PR such that angle PQS= angle RQS. prove that S is equidistant from PQ and QR

Answer 1

Given : seg PS ≅ seg PT                    … (1)

 seg SQ ≅ seg TR                              … (2)                    

In △PST,

seg PS ≅ seg PT

⇒∠S = ∠T                … (isosceles triangle theorem)

In △PST, by angle-sum property,

∠P +∠S + ∠T = 180°

⇒∠P = 180°-∠S -∠T                    … (3)

 Adding (1) and (2), we get:

 seg PS +seg SQ ≅seg PT + seg TR

⇒ seg PQ ≅ seg PR

Now, in △PQR,

⇒∠Q = ∠R                … (isosceles triangle theorem)

In △PQR, by angle-sum property,

∠P +∠Q + ∠R = 180°

⇒∠P = 180°-∠Q -∠R                    … (4)

From (3) and (4),

180°-  ∠Q -∠R = 180° -∠S -∠T

⇒∠Q + ∠R = ∠S + ∠T

But ∠Q = ∠R and ∠S = ∠T

⇒ 2 ∠Q = 2 ∠S

⇒∠Q = ∠S

⇒∠Q = ∠S and ∠T = ∠R

But these angles are corresponding angles formed by transversals PQ and PR, respectively.

So side ST is parallel to side QR.

Answer 2

Answer:

Given : seg PS ≅ seg PT                    … (1)

seg SQ ≅ seg TR                              … (2)                    

In △PST,

seg PS ≅ seg PT

⇒∠S = ∠T                … (isosceles triangle theorem)

In △PST, by angle-sum property,

∠P +∠S + ∠T = 180°

⇒∠P = 180°-∠S -∠T                    … (3)

Adding (1) and (2), we get:

seg PS +seg SQ ≅seg PT + seg TR

⇒ seg PQ ≅ seg PR

Now, in △PQR,

⇒∠Q = ∠R                … (isosceles triangle theorem)

In △PQR, by angle-sum property,

∠P +∠Q + ∠R = 180°

⇒∠P = 180°-∠Q -∠R                    … (4)

From (3) and (4),

180°-  ∠Q -∠R = 180° -∠S -∠T

⇒∠Q + ∠R = ∠S + ∠T

But ∠Q = ∠R and ∠S = ∠T

⇒ 2 ∠Q = 2 ∠S

⇒∠Q = ∠S

⇒∠Q = ∠S and ∠T = ∠R

But these angles are corresponding angles formed by transversals PQ and PR, respectively.

So side ST is parallel to side QR.