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.
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.