In triangle ABC; $∠$ A = $60^{0}$ , $∠$ C = $40^{0}$ and bisector of angle ABC meets side AC at point P. Show that BP= CP.

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Solution

In $△$ ABC

B= $180^{0} - 60^{0} - 40^{0}$

B= $80^{0}$

Now $∠ A B P = ∠ P B C$ {given}

Also BP=CP {given}

Hence PB=PC

$∴ ∠ P B C = ∠ P C B = 40^{0}$

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In triangle ABC; ∠ A = 600, ∠ C = 400 and bisector of angle ABC meets side AC at point P. Show that BP= CP.

In △ ABC B= 1800−600−400 B= 800 Now ∠ABP=∠PBC {given} Also BP=CP {given} Hence PB=PC ∴∠PBC=∠PCB=400

In triangle ABC; $∠$ A = $60^{0}$ , $∠$ C = $40^{0}$ and bisector of angle ABC meets side AC at point P. Show that BP= CP.

In $△$ ABC

B= $180^{0} - 60^{0} - 40^{0}$

B= $80^{0}$

Now $∠ A B P = ∠ P B C$ {given}

Also BP=CP {given}

Hence PB=PC

$∴ ∠ P B C = ∠ P C B = 40^{0}$