# Plzz solve1,2.... URGENT Dear Student,

It is given in the question that
As,

Thus replacing the values in equation (i), we get,

Thus, considering $△GCB$ and $△DCE$, we can see that,

Hence, based on ASA (Angle Side Angle) property we can say that $△GCB$ and $△DCE$ are congruent with each other.

In right $△OLP$, we can see that,

And, In right $△OMP$, we can see that,

Thus, from (i) and (ii), we can see that,

Considering $△OLP$ and $△OMP$, we can see that;

Hence, based on SAS (Side Angle Side) property we can say that $△OLP$ and $△OMP$ are congruent with each other.

Therefore,
Since,

Thus, we can definitely say that $OP$ is the bisector of $\angle AOB$.

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Hope.its useful is it ??
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HELLO:
HERE IS THE SOLUTION FOR 1ST QUESTION:
GIVEN:ANGLE ABG = ANGLE DEF
BC =CE
TO PROVE THAT TRIANGLE GCB IS CONGRUENT TO TRIANGLE DFE
PROOF: IN TRIANGLE GCB AND TRIANGLE DFE:
AS ANGLE ABG = ANGLE DEF ,
THEN: 1800- ANGLE ABG=1800 - ANGLE DEF
SO :ANGLE GBC = ANGLE DCE
BC=CE (GIVEN)
ANGLEGCB= ANGLE DCE(VERTICALLY OPPOSITE ANGLE)
THEREFORE, TRIANGLE GCB IS CONGRUENT TO TTRIANGLE DCE (BY ASA)

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HERE IS THE ANSWER FOR UR SECOND QUESTION:
GIVEN; PL=PM AND PM AND PL IS PERPENDICULAR TO BO AND AO
TO PROVE ; OP IS THE BISECTOR OF ANGLE AOB
PROOF:
IN TRIANGLE PLO:
h2= p2+b2

OP2=PL2+OL2

OP2 = PM+OL(AS PL=PM)
OL= OP- PM(BUT WE ALSO KNOW THAT MO= OP2 - PM2)
SO OL = PM

IN TRIANLE PLO AND TRIANGLE PMO:
OL = OM(PROVED ABOVE)
PL = PM (GIVEN)
OP=OP (COMMON)
THEREFORE TRIANGLE POL IS CONGRUENT TO TRIANGLE POM (BY SSS)
THEREFORE ANGLE LOP = ANGLE MOP (BY CPCT)
SO OP BISECT ANGLE AOB (PROVED).
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