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; ∠ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Mathematics

Co-interior Angles Theorem

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.

Open in App

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

Suggest Corrections

Similar questions

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 triangle ABC, angle B = 35^o, angle C = 65^o and the bisector of angle BAC meets BC in P. Choose the correct descending order of AP, BP and CP

In triangle ABC; angle ABC = $90^{0}$ and P is a point on AC such that $∠$ PBC = $∠$ PCB. Show that : BP = CP.

In $Δ$ ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that:
(i) CB: BA =CP: PA
(ii)AB $\times$ BC = BP $\times$ CA

Q. In

△ A B C

, angle

A B C

is equal to twice the angle

A C B

, and bisector of angle

A B C

meets the opposite side at point

P

, show that:
(i)

C B : B A = C P : P A

(ii)

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

Join BYJU'S Learning Program

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