Question

In the figure given below, angle x= angle y and PQ=QR. Prove that PE=RS

Answer 1

Solution:-
∠ PSR = x
∠ QSR + ∠ PSR = 180° (Linear pair)
∠ QSR = 180° - x
∠ REP = y
∠ QEP + ∠ REP = 180°
∠ QEP = 180° - y = 180° - x (Since x = y)
In Δ QEP and Δ QSR
∠ QEP = ∠ QSR (Proved above)
∠ PQE = ∠ RQS (Common)
PQ = QR (Given)
By AAS congruence
Δ QEP ≡ Δ QSR
So,
PE = RS (CPCT)

Answer 2

Solution:-

∠ PSR = x

∠ QSR + ∠ PSR = 180° (Linear pair)

∠ QSR = 180° - x

∠ REP = y

∠ QEP + ∠ REP = 180°

∠ QEP = 180° - y = 180° - x (Since x = y)

In Δ QEP and Δ QSR

∠ QEP = ∠ QSR (Proved above)

∠ PQE = ∠ RQS (Common)

PQ = QR (Given)

By AAS congruence

Δ QEP ≡ Δ QSR

So,

PE = RS (CPCT)