Question

please help me for the question​

Answer 1

Step-by-step explanation:

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Answer 2

Answer: LET ∠PQS AND ∠RQS=Y

SINCE PQ║RS AND QS IS A TRANSVERSAL

⇒∠PQS=∠QSR  (ALT INT ANGLES)

ALSO

∠PQS=∠RQS  (GIVEN)

SO,

∠QSR=∠RQS

IN Δ RQS

∠QSR=∠RQS

⇒QR=RS   (SIDES OPP TO EQUAL ANGLES ARE EQUAL)

SINCE PQ║RS AND PR IS A TRANSVERSAL

⇒∠QPR=X  (ALT INT ANGLES)

IN Δ PQR

∠PQR+∠QPR= ∠PRT  (EXTERIOR ANGLE PROPERTY)

∠PQS+∠RQS+X= ∠PRS+∠TRS

Y+Y+X=X+X

2Y= 2X-X= X

Y= 1/2 X

⇒∠PQS= 1/2 ∠QPR  (SUBSTITUTING VALUES)

HENCE PROVED:)

Step-by-step explanation:

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JUST TOO LONG

AND MY FINGERS HURT BY TYPING SO MUCH