filmov
tv
∠B=2∠C. BC such that AD bisects ∠BAC and AB=CD. BE is the bisector of ∠B. ∠BAC = 72°
Показать описание
In the figure, ABC is a triangle in which ∠B=2∠C. D is a point on side BC such that AD bisects ∠BAC and AB=CD. BE is the bisector of ∠B. The measure of ∠BAC is 72°
*solution*
The correct option is A
72∘
In the figure, ∠B=2∠C, AD and BE are the bisectors of ∠A and ∠B respectively, AB=CD
Let ∠C=x, then ∠B=2x and
Let ∠A=2y
∴ ∠ABE=∠CBE=x
and ∠BAD=∠CAD=y
In ΔBCE,
∴ ∠EBC=∠BCE=x
∴ BE=EC
In ΔABE and ΔDCE,
AB=DC (Given)
∠ ABE=∠C (Each=x)
BE=EC (Proved)
∴ ΔABE≅ ΔDCE (SAS axiom)
∴ ∠BAE=∠EDC=2y
and AE=ED
∴ ∠EAD=∠EDA=y
In ΔABD,
Ext. ∠ADC=2x+y=2y+y
⇒ 2x=2y⇒x=y
Now in ΔABC,
∠A+∠B+∠C=180∘
⇒ 2y+2x+x=180∘⇒ 2x+2x+x=180∘
⇒ 5x=180∘⇒ x=180∘5=36∘
∴ ∠BAC=2y=2x=2×36∘=72∘
#mathematics
#9th
#cbse
#rdsharma
#triangle
*solution*
The correct option is A
72∘
In the figure, ∠B=2∠C, AD and BE are the bisectors of ∠A and ∠B respectively, AB=CD
Let ∠C=x, then ∠B=2x and
Let ∠A=2y
∴ ∠ABE=∠CBE=x
and ∠BAD=∠CAD=y
In ΔBCE,
∴ ∠EBC=∠BCE=x
∴ BE=EC
In ΔABE and ΔDCE,
AB=DC (Given)
∠ ABE=∠C (Each=x)
BE=EC (Proved)
∴ ΔABE≅ ΔDCE (SAS axiom)
∴ ∠BAE=∠EDC=2y
and AE=ED
∴ ∠EAD=∠EDA=y
In ΔABD,
Ext. ∠ADC=2x+y=2y+y
⇒ 2x=2y⇒x=y
Now in ΔABC,
∠A+∠B+∠C=180∘
⇒ 2y+2x+x=180∘⇒ 2x+2x+x=180∘
⇒ 5x=180∘⇒ x=180∘5=36∘
∴ ∠BAC=2y=2x=2×36∘=72∘
#mathematics
#9th
#cbse
#rdsharma
#triangle
0:07:31
∠B=2∠C. BC such that AD bisects ∠BAC and AB=CD. BE is the bisector of ∠B. ∠BAC = 72°
0:09:16
A B C is a triangle in which ∠ B=2 ∠ C. D is a point on B C such that A D bisects ∠ B A C and A B......
0:07:48
ABC is a triangle in which B=2C, D is a point on side BC such that AD bisects BAC and AB=CD. BE is t
0:05:53
|| Most Important Question || ABC is a triangle in which angle B = 2C . D is a point on BC ||
0:02:10
D and E are respectively the points on the sides AB and AC of a triangle ABC such that AD= 2 cm, BD
0:06:06
ABC is a triangle in which `angleB=2angleC`. D is a point on BC such that AD bisects `ang
0:05:21
20.In a ∆ABC,AD perp. to BC such that AD^2 = BD.CD. Prove that ∆ABC is right-angled at A.
0:09:30
In the triangle ABC ,DE parallel to BC,So that AD=2.4cm AE=3.2cm and EC=4.8Cm and then AB=
0:06:20
If A B C is an equilateral triangle such that A D ⊥ B C, then A D^2= (a) 3/2 D C^2 (b) 2 D C^2 (c......
0:22:32
ABC is a triangle in which ∠B = 2∠C. D is a point on BC such that AD bisects ∠BAC and AB = CD. Prove...
0:05:45
In a triangle ABC, D and E are point on BC such that AD = AE and ∠BAD = ∠CAE. If AB = (2p + 3)..
0:11:10
In ΔABC, D⊥BC , such that AD2 = BD×CD. Prove that ΔABC is right angled at
0:02:23
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC | Exercise 6.3 class 10 Q13
0:01:16
ABC is an equilateral triangle P is a point on BC such that BP:PC=2:1. Prove that : 9AP^2 = 7AB^2
0:18:50
In triangle ABC, AD is perpendicular to BC such that AC^2 = BD.CD . Prove that triangle ABC is
0:10:16
In a triangle ABC, angle B is twice angle C. AD bisects angle BAC. AB = DC. Find angle BAC.
0:03:01
In triangle ABC , angle B is equal to twice of angle C , D is a point on BC such that AD bisect...
0:00:23
Scientist Modi Ji | Modi ji ka mathematics | Extra 2ab | Prem Sir
0:04:30
ABC is a right triangle right−angled at C. Let BC=a, CA=b, AB=c and p be the length of perpendicular...
0:00:57
Quadratic Equations IIT Questions No 05 ( X Class)
0:10:43
If ABC is a right angled triangle such that AB=AC and bisector of angle C intersect the side AB at D...
0:05:43
In triangle ABC, AD is perpendicular to BC, AD^2=BD×CD. Prove that triangle is a right triangle.
0:03:22
Sibibala - B2C ft King Saha Official Video
0:13:56
D is a point on the side BC of a ∆ ABC such that AD bisects ∠BAC. Then (A) BD = CD (B
Комментарии