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good afternoon_____________❤️❤️❤️
_____________________
Given: OA = OB and OP = OQ
To prove:
(i) PX = QX
(ii) AX = BX
Proof: In OAQ and OPB, we have,
OA = OB [Given]
O = O [Common]
OQ = OP [Given]
Thus by Side-Angle-Side criterion of congruence, we have
OAQ OPB
The corresponding parts of the congruent triangles are equal.
Therefore, OBP = OAQ ——— (1)
Thus, in BXQ and PXA, we have
BQ = OB – OQ
and, PA = OA – OP
But, OP = OQ
and OA =OB [Given]
Therefore, we have, BQ = PA ———- (2)
Now consider triangles BXQ and PXA.
BXQ = PXA [vertically opposite angles]
OBP = OAQ [from (1)]
BQ = PA [from (2)]
Thus by Angle-Angle-Side criterion of congruence, we have,
Therefore, BXQ PPXA
PX = QX [c.p.c.t]
AX = BX [c.p.c.t]
good afternoon_____________❤️❤️❤️
_____________________
Given: OA = OB and OP = OQ
To prove:
(i) PX = QX
(ii) AX = BX
Proof: In OAQ and OPB, we have,
OA = OB [Given]
O = O [Common]
OQ = OP [Given]
Thus by Side-Angle-Side criterion of congruence, we have
OAQ OPB
The corresponding parts of the congruent triangles are equal.
Therefore, OBP = OAQ ——— (1)
Thus, in BXQ and PXA, we have
BQ = OB – OQ
and, PA = OA – OP
But, OP = OQ
and OA =OB [Given]
Therefore, we have, BQ = PA ———- (2)
Now consider triangles BXQ and PXA.
BXQ = PXA [vertically opposite angles]
OBP = OAQ [from (1)]
BQ = PA [from (2)]
Thus by Angle-Angle-Side criterion of congruence, we have,
Therefore, BXQ PPXA
PX = QX [c.p.c.t]
AX = BX [c.p.c.t]
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