In ∆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.

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Solution

In triangle BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
$\frac{Q R}{P R} = \frac{B Q}{B P}$
$\Rightarrow B P \times Q R = B Q \times P R$
This completes the proof.

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In ∆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 triangle ​BQP, BR bisects angle B. Applying angle bisector theorem, we get: QRPR = BQBP ⇒ BP × QR = BQ × PR This completes the proof.

In ∆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 triangle BQP, BR bisects angle B.
Applying angle bisector theorem, we get:
$\frac{Q R}{P R} = \frac{B Q}{B P}$
$\Rightarrow B P \times Q R = B Q \times P R$
This completes the proof.