Advertisements
Advertisements
प्रश्न
In ΔABC, AB = AC and the bisectors of angles B and C intersect at point O.
Prove that : (i) BO = CO
(ii) AO bisects angle BAC.
Advertisements
उत्तर
In ΔABC,
AB = AC
⇒ ∠B = ∠C ...( angles opposite to equal sides are equal )
⇒ `1/2 ∠"B" = 1/2∠"C"`
⇒ ∠OBC = ∠OCB ...[ ∵ OB and OC are bisectors of ∠B and ∠C respectively, ∠OBC = `1/2∠"B" and ∠"OCB" = 1/2∠"C"` ] ...(i)
⇒ OB = OC ...( Sides opposite to equal angles are equal ) ...(ii)
Now, in ΔABO and ΔACO,
AB = AC ...( given )
∠OBC = ∠OCB ...[ from(i) ]
OB = OC ...[ from(ii) ] ...( proved )
∴ ΔABO ≅ ΔACO ...( by SAS congruence criterion )
⇒ ∠BAO = ∠CAO ...( c.p.c.t. )
⇒ AO bisects ∠BAC ...(proved)
APPEARS IN
संबंधित प्रश्न
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see the given figure). Show that:
- ΔAMC ≅ ΔBMD
- ∠DBC is a right angle.
- ΔDBC ≅ ΔACB
- CM = `1/2` AB
Which congruence criterion do you use in the following?
Given: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ΔLMN ≅ ΔGFH
In ΔABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ΔPQR, ∠P = 30°, ∠Q = 40° and ∠R = 110°
A student says that ΔABC ≅ ΔPQR by AAA congruence criterion. Is he justified? Why or why not?
Prove that the perimeter of a triangle is greater than the sum of its altitudes.
In the given figure, prove that:
CD + DA + AB + BC > 2AC
ABC is an isosceles triangle in which AB = AC. BE and CF are its two medians. Show that BE = CF.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB = CE.
A triangle ABC has ∠B = ∠C.
Prove that: The perpendiculars from the mid-point of BC to AB and AC are equal.
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
