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Question

BO and CO are respectively the bisectors of B and C of ΔABC. AO produced meets BC at P.


Show that [4 MARKS]

(i) ABBP=AOOP

(ii) ACCP=AOOP

(iii) ABAC=BPPC

(iv) AP is the bisector of BAC.


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Solution

Each proof : 1 Mark each



(i) In ΔABP, BO is the bisector of B

ABBP=AOOP [Angle bisector theorem]

(ii) In ΔACP, OC is the bisector of C.

ACCP=AOOP [Angle bisector theorem]

(iii) We have, proved that

ABBP=AOOP and ACCP=AOOP

ABBP=ACCP

ABAC=BPPC

(iv) As proved above that in ΔABC, we have

ABAC=BPPC

AP is the bisector of BAC.


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