In $Δ A B C$ , the bisector of $∠ B$ meets AC at D. A line PQ || AC meets AB, BC and BD at P, Q and R respectively.

Show that PR $\times$ BQ = QR $\times$ BP.

Open in App

Solution

In tringle BRQ and triangle BRP,

angle B is an common angle....1

angle R is also an common angle.......2

therefore,triangle BRQ is proportion to triangle BRP (AA congruence)

therefore,

BQ/BP=QR/RP

PRBQ=QRBP

Hence Proved..

Suggest Corrections

Similar questions

Q. In

△ A B C

, the bisector of

∠ B

meets AC at D. A line PQ||AC meets AB,BC and BD at P,Q and R respectively. Show that
(i)

P R . B Q = Q R . B P

(ii)

A B \times C Q = B C \times A P

BO and CO are respectively the bisectors of $∠ B$ and $∠ C$ of $Δ A B C$ . AO produced meets BC at P.

Show that [4 MARKS]

(i) $\frac{A B}{B P} = \frac{A O}{O P}$

(ii) $\frac{A C}{C P} = \frac{A O}{O P}$

(iii) $\frac{A B}{A C} = \frac{B P}{P C}$

(iv) $A P$ is the bisector of $∠ B A C$ .

BO and CO are respectively the bisectors of angle B and C of ∆ABC. AO produced meets BC at P. Show that AO/BP = AO/OP , AC/CP = AO/OP ,

AB/AC = BP/PC , AP is the bisector of angle BAC.

Q. ∆ABC and ∆DBC lie on the same side of BC, as shown in the figure. From a point P on BC, PQ ∥ AB and PR ∥ BD are drawn, meeting AC at Q and CD at R, respectively. Prove that QR ∥ AD.

$Δ A B C$ and $Δ D B C$ lie on the same side of BC, as shown in the figure. From a point P on BC, PQ || AB and PR || BD are drawn, meeting AC at Q and CD at R respectively. Prove that QR || AD.

Join BYJU'S Learning Program

In Δ ABC, the bisector of ∠B meets AC at D. A line PQ || AC meets AB, BC and BD at P, Q and R respectively. Show that PR × BQ = QR × BP.

In tringle BRQ and triangle BRP, angle B is an common angle....1 angle R is also an common angle.......2 therefore,triangle BRQ is proportion to triangle BRP (AA congruence) therefore, BQ/BP=QR/RP PR*BQ=QR*BP Hence Proved..

In $Δ A B C$ , the bisector of $∠ B$ meets AC at D. A line PQ || AC meets AB, BC and BD at P, Q and R respectively.

Show that PR $\times$ BQ = QR $\times$ BP.

In tringle BRQ and triangle BRP,

angle B is an common angle....1

angle R is also an common angle.......2

therefore,triangle BRQ is proportion to triangle BRP (AA congruence)

therefore,

BQ/BP=QR/RP

PRBQ=QRBP

Hence Proved..