In the figure, seg PQ || side BC and seg QR || side AB then (i) Find AQ/QC (ii) What would be CR/RB and hence find BR/RC (iii) Is BP/PA = BR/RC ?
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Answered by
13
Answer:
In ΔAPQ & ΔABC
∠BAC=∠PAQ (same angle)
∠ABC=∠APQ (PQ∥BC)
∠AQP=∠ACB (PQ∥BC)
Triangles are similar
AQ
AC
=
AP
AB
=
3
5
AQ
AC
−1=
AQ
AC−AQ
=
3
5
−1=
3
2
⇒
AQ
QC
=
3
2
,
QC
AQ
=
2
3
(ii) ΔCQR and ΔCAB
∠ACB=∠QCR (same angle)
∠CQR=∠CAB (QR∥AB)
∠CRQ=∠CBA (QR∥AB)
Triangles are similar
CR
RB
+1=
CR
RB+CR
=
CR
CB
CR
CB
=
CQ
CA
=
2
5
CR
CB
=
2
5
CR
RB
=
2
5
−1
BR/RC=2/3
Answered by
8
Answer:
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