Question

In fig. 3.20 seg PS $\cong$ seg PT, seg SQ $\cong$ seg TR, then show that Side ST $\parallel$ Side QR

Answer 1

Given : seg PS $\cong$ seg PT … (1)
seg SQ $\cong$ seg TR … (2)
In $△$PST,
seg PS $\cong$ seg PT
$\Rightarrow$$\angle$S = $\angle$T … (isosceles triangle theorem)
In $△$PST, by angle-sum property,
$\angle$P +$\angle$S + $\angle$T = 180$°$
$\Rightarrow$ $\angle$P = 180$°$$-$ $\angle$S $-$ $\angle$T … (3)

Adding (1) and (2), we get:
seg PS +seg SQ $\cong$seg PT + seg TR
$\Rightarrow$ seg PQ $\cong$ seg PR

Now, in $△$PQR,
$\Rightarrow$$\angle$Q = $\angle$R … (isosceles triangle theorem)
In $△$PQR, by angle-sum property,
$\angle$P +$\angle$Q + $\angle$R = 180$°$
$\Rightarrow$$\angle$P = 180$°$ $-$$\angle$Q $-$ $\angle$R … (4)
From (3) and (4),
180$°$ $-$ $\angle$Q $-$ $\angle$R = 180$°$ $-$ $\angle$S $-$$\angle$T
$\Rightarrow$ $\angle$Q + $\angle$R = $\angle$S + $\angle$T
But $\angle$Q = $\angle$R and $\angle$S = $\angle$T
$\Rightarrow$ 2 $\angle$Q = 2 $\angle$S
$\Rightarrow$$\angle$Q = $\angle$S
$\Rightarrow$$\angle$Q = $\angle$S and $\angle$T = $\angle$R
But these angles are corresponding angles formed by transversals PQ and PR, respectively.
So side ST is parallel to side QR.