DE parallel to QR and AP and BP are bisector of angle EAB and angle RBA, respectively . Find angle APB
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IN TRIANGLE PQR
AC||PR (GIVEN)
HENCE
\frac{qc}{cp} = \frac{qa}{ar}
cp
qc
=
ar
qa
(by BPT THEOREAM)
hence
\frac{15}{cp } = \frac{12}{20}
cp
15
=
20
12
NOW ...
CP =
\frac{15}{12} \times 20
12
15
×20
hence
CP = 25cm -----------eq 1
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NOW IN TRIANGLE PQR AGAIN.....
CB || QR ( GIVEN )
HENCE
\frac{pc}{qc} = \frac{pb}{br}
qc
pc
=
br
pb
( BY BPT THEOREAM )
HENCE
from eq 1 value of PC = 25 cm
\frac{25}{15} = \frac{15}{br}
15
25
=
br
15
HENCE BR =
\frac{15}{25} \times 15
25
15
×15
hence
BR = 9 cm
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