Advertisements
Advertisements
प्रश्न
A point O is taken inside a rhombus ABCD such that its distance from the vertices B and D are equal. Show that AOC is a straight line.
Advertisements
उत्तर
In ΔAOD and ΔAOB,
AD = AB ...(given)
AO = AO ...(Common)
OD = OB ...(given)
⇒ ΔAOD ≅ ΔAOB ...(by SSS congruence criterion)
⇒ ∠AOD = ∠AOB ...(c.p.c.t.) ...(i)
Similarly, ΔDOC ≅ ΔBOC
⇒ ∠DOC = ∠BOC ...(c.p.c.t.) ...(ii)
But, ∠AOB + ∠AOD + ∠COD + ∠BOC = 4 Right angles ...[ Sum of the angles at a point is 4 Right angles ]
⇒ 2∠AOD + 2∠COD = 4 Right angles ....[ Using (i) and (ii) ]
⇒ ∠AOD + ∠COD = 2 Right angles
⇒ ∠AOD + ∠COD = 180°
⇒ ∠AOD and ∠COD form a linear pair.
⇒ AO and OC are in the same straight line.
⇒ AOC is a straight line.
APPEARS IN
संबंधित प्रश्न
AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (See the given figure). Show that
- ΔDAP ≅ ΔEBP
- AD = BE
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?
If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.
In a triangle ABC, D is mid-point of BC; AD is produced up to E so that DE = AD. Prove that:
AB is parallel to EC.
From the given diagram, in which ABCD is a parallelogram, ABL is a line segment and E is mid-point of BC.
Prove that:
(i) ΔDCE ≅ ΔLBE
(ii) AB = BL.
(iii) AL = 2DC
In the following figure, BL = CM.
Prove that AD is a median of triangle ABC.
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that: BD = CD
In the following figure, AB = AC and AD is perpendicular to BC. BE bisects angle B and EF is perpendicular to AB.
Prove that : ED = EF
In the parallelogram ABCD, the angles A and C are obtuse. Points X and Y are taken on the diagonal BD such that the angles XAD and YCB are right angles.
Prove that: XA = YC.
In quadrilateral ABCD, AD = BC and BD = CA.
Prove that:
(i) ∠ADB = ∠BCA
(ii) ∠DAB = ∠CBA
