English

In δ Abc, ∠B = 35°, ∠C = 65° and the Bisector of ∠Bac Meets Bc in P. Arrange Ap, Bp and Cp in Descending Order. - Mathematics

Advertisements
Advertisements

Question

In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.

Answer in Brief
Advertisements

Solution

It is given that

∠B = 35°

∠C = 65°

AP is the bisector of ∠CAB

We have to arrangeAP, BPand CPin descending order.

In  ΔACP we have

∠ACP = 65°

∠CAP = 40°(As AP is the bisector of ∠CAB

So  AP > CP (Sides in front or greater angle will be greater)              ........(1)

In  ΔABP we have

∠BAP = 40°(As AP is the bisector of ∠CAB)

Since, 

∠BAP >∠ABP

So  BP > AP                ..........(2)

Hence 

From (1) & (2) we have

 BP > AP > CP

shaalaa.com
Criteria for Congruence of Triangles
  Is there an error in this question or solution?
Chapter 12: Congruent Triangles - Exercise 12.6 [Page 81]

APPEARS IN

RD Sharma Mathematics [English] Class 9
Chapter 12 Congruent Triangles
Exercise 12.6 | Q 5 | Page 81

RELATED QUESTIONS

In quadrilateral ACBD, AC = AD and AB bisects ∠A (See the given figure). Show that ΔABC ≅ ΔABD. What can you say about BC and BD?


Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:

  1. ΔAPB ≅ ΔAQB
  2. BP = BQ or B is equidistant from the arms of ∠A.


In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.


If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.  


Which of the following statements are true (T) and which are false (F):

Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.


In the given figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is


The perpendicular bisectors of the sides of a triangle ABC meet at I.

Prove that: IA = IB = IC.


In ∆ABC, AB = AC. Show that the altitude AD is median also.


ABCD is a parallelogram. The sides AB and AD are produced to E and F respectively, such produced to E and F respectively, such that AB = BE and AD = DF.

Prove that: ΔBEC ≅ ΔDCF.


In the following figure, ABC is an equilateral triangle in which QP is parallel to AC. Side AC is produced up to point R so that CR = BP.

Prove that QR bisects PC.
Hint: ( Show that ∆ QBP is equilateral
⇒ BP = PQ, but BP = CR
⇒ PQ = CR ⇒ ∆ QPM ≅ ∆ RCM ).


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×