In triangle ABC- - A - 60- - C - 40- and bisector of angle ABC meets side AC at point P- Check whether BP - CP-

In triangle ABC ∠ A+∠ ABC+∠ C=180 #176; #160; ∠ ABC=180 #176; 60 #176; 40 #176;=80 #176; ∠ PBC=80 #176;/2=40 #176; #160; ∠ ...

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Standard IX

Mathematics

Every Point on the Bisector of an Angle Is Equidistant from the Sides of the Angle.

In triangle A...

Question

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

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Solution

In triangle ABC
$∠ A + ∠ A B C + ∠ C = 180 °$
$∠ A B C = 180 ° - 60 ° - 40 ° = 80 °$
$∠ P B C = 80 ° / 2 = 40 °$
$∠ C = 40 °$
So, triangle PCB is Isosceles Triangle. So, Corresponding side will be equal.
So, BP=CP
Answer 1

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In triangle ABC; ∠A=60∘, ∠C=40∘ and bisector of angle ABC meets side AC at point P. Check whether BP = CP.

In triangle ABC∠A+∠ABC+∠C=180° ∠ABC=180°−60°−40°=80°∠PBC=80°/2=40° ∠C=40°So, triangle PCB is Isosceles Triangle. So, Corresponding side will be equal.So, BP=CPAnswer 1

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

In triangle ABC
$∠ A + ∠ A B C + ∠ C = 180 °$
$∠ A B C = 180 ° - 60 ° - 40 ° = 80 °$
$∠ P B C = 80 ° / 2 = 40 °$
$∠ C = 40 °$
So, triangle PCB is Isosceles Triangle. So, Corresponding side will be equal.
So, BP=CP
Answer 1