Question

plss help....



i willl mark as brainlist to the best answer....​

Answer 1

(i) In ∆ QRS and ∆QPS

QR = QP ( given )

QS = QS ( common )

$\sf{\angle{PQS} = \angle{RQS}}$ (As QS is angle bisector of angle Q )

∵ ∆ QRS ≅ ∆QPS

(ii) The three pairs are :-

QR = QP ( given )

QS = QS ( common )

$\sf{\angle{PQS} = \angle{RQS}}$

Answer 2

Answer:

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