In Fig 3.20 Arcs drawn with some radius and centre B, intersect the sides of $∠$ ABC at point P and Q respectively. Again arcs are drawn with centre P and Q, keeping the radius constant. These arcs intersect at R. Show that the ray BR bisects $∠$ ABC.

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Solution

$Given : All arcs are drawn with constant radius . Construction : Join PR, QR and BR . In △ BPR and △ BQR, BP = BQ (arc drawn with constant radius) PR = QR (arc drawn with constant radius) BR = BR (common) ∴ △ BPR ≅ △ BQR (SSS test) ∠ ABR = ∠ CBR (by c . a . c . t) Hence, ray BR bisects ∠ ABC .$

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In Fig 3.20 Arcs drawn with some radius and centre B, intersect the sides of ∠ABC at point P and Q respectively. Again arcs are drawn with centre P and Q, keeping the radius constant. These arcs intersect at R. Show that the ray BR bisects ∠ABC.

Given : All arcs are drawn with constant radius.Construction: Join PR, QR and BR.In △BPR and △BQR,BP = BQ (arc drawn with constant radius)PR = QR (arc drawn with constant radius)BR =BR (common) ∴△BPR ≅△BQR (SSS test)∠ABR = ∠CBR (by c.a.c.t) Hence, ray BR bisects ∠ABC.

In Fig 3.20 Arcs drawn with some radius and centre B, intersect the sides of $∠$ ABC at point P and Q respectively. Again arcs are drawn with centre P and Q, keeping the radius constant. These arcs intersect at R. Show that the ray BR bisects $∠$ ABC.