Answer:
Correct option is
b. If two sides and one included angle are equal in triangles PQS and PRS, then their corresponding sides are also equal.
Step-by-step explanation:
Here, we are given the line RQ, which is divided in two equal parts by a line PS which is perpendicular to RQ.
The foot S of PS is on the line RQ.
First of all, let us do a construction here.
Join the point R with P and P with Q.
Please refer to the attached image.
Now, let us consider the triangles PQS and PRS:
- [tex]\angle PSR = \angle PSQ = 90^\circ[/tex]
- Side PS = PS (Common side in both the triangles)
Now, Two sides and the angle included between the two triangles are equal.
So by SAS congruence we can say that [tex]\triangle PRS \cong \triangle PQS[/tex]
Therefore, the corresponding sides will also be equal.
RP = QP
RP is the distance between R and P.
QP is the distance between Q and P.
Hence, to prove that P is equidistant from R and Q, we have proved that:
b. If two sides and one included angle are equal in triangles PQS and PRS, then their corresponding sides are also equal.
