Advertisements
Advertisements
Question
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
Solution
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
RELATED QUESTIONS
In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Which congruence criterion do you use in the following?
Given: AC = DF
AB = DE
BC = EF
So, ΔABC ≅ ΔDEF
You want to show that ΔART ≅ ΔPEN,
If it is given that AT = PN and you are to use ASA criterion, you need to have
1) ?
2) ?
In two triangles ABC and ADC, if AB = AD and BC = CD. Are they congruent?
If the following pair of the triangle is congruent? state the condition of congruency :
In Δ ABC and Δ DEF, AB = DE, BC = EF and ∠ B = ∠ E.
If the following pair of the triangle is congruent? state the condition of congruency:
In ΔABC and ΔPQR, AB = PQ, AC = PR, and BC = QR.
The perpendicular bisectors of the sides of a triangle ABC meet at I.
Prove that: IA = IB = IC.
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 following diagram, ABCD is a square and APB is an equilateral triangle.
- Prove that: ΔAPD ≅ ΔBPC
- Find the angles of ΔDPC.
In the following diagram, AP and BQ are equal and parallel to each other. 
Prove that: AB and PQ bisect each other.
