Question

Please help me in solving this question

Answer 1

hiiii dear___________

good afternoon___________❤️❤️❤️

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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]

Answer 2

OA= OB
OP=OQ

To prove i) PX=QX
ii) AX=BX


In BOP & AOQ
OP=OQ (given)
OA=OB (given)
angle BOP=AOQ (common)

By SAS property,
ΔBOP = ΔAOQ
angle A=B (CPCT)

In ΔBXQ & ΔAXP,
BQ=AP (OA-OP = OB-OQ)
angle BXQ=AXP (VOA)
angle A=B (proven)

By ASA property,
ΔBXQ = ΔAXP

PX=QX & AX=BX (CPCT)
proven√