Given,
AB and CD are chords intersecting at E.
AB = CD
To prove,
AE = DE and CE = BE
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

Proof,
OM bisects AB (OM ⊥ AB)
ON bisects CD (ON ⊥ CD)
As AB = CD thus,
AM = ND --- (i)
and MB = CN --- (ii)
In ΔOME and ΔONE,
∠OME = ∠ONE (Perpendiculars)
OE = OE (Common)
OM = ON (AB = CD and thus equidistant from the centre)
ΔOME ≅ ΔONE by RHS congruence condition.
ME = EN by CPCT --- (iii)
From (i) and (ii) we get,
AM + ME = ND + EN
⇒ AE = ED
From (ii)  and (iii) we get,
MB - ME = CN - EN
⇒ EB = CE

Given,
AB and CD are chords intersecting at E.
AB = CD, PQ is the diameter.
To prove,
∠BEQ = ∠CEQ
Construction,
OM ⊥ AB and ON ⊥ CD. OE is joined.

In ΔOEM and ΔOEN,
OM = ON (Equal chords are equidistant from the centre)
OE = OE (Common)
∠OME = ∠ONE (Perpendicular)
ΔOEM ≅ ΔOEN by RHS congruence condition.
Thus,
∠MEO = ∠NEO by CPCT
⇒ ∠BEQ = ∠CEQ

If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig. 10.25).

Answer

OM ⊥ AD is drawn from O.
OM bisects AD as OM ⊥ AD.
⇒ AM = MD --- (i)
also, OM bisects BC as OM ⊥ BC.
⇒ BM = MC --- (ii)
From (i) and (ii),
AM - BM = MD - MC
⇒ AB = CD

 

Let A, B and C represent the positions of Reshma, Salma and Mandip respectively.
AB = 6cm and BC = 6cm.
Radius OA = 5cm
BM ⊥ AC is drawn.
ABC is an isosceles triangle as AB = BC, M is mid-point of AC. BM is perpendicular bisector of AC and thus it passes through the centre of the circle.
Let AM = y and OM = x then BM = (5-x).
Applying Pythagoras theorem in ΔOAM,
OA² = OM² + AM² 
⇒ 5² = x² + y² --- (i)
Applying Pythagoras theorem in ΔAMB,
AB² = BM² + AM² 
⇒ 6² = (5-x)² + y² --- (ii)
Subtracting (i) from (ii), we get
36 - 25 = (5-x)² - x² -
⇒ 11 = 25 - 10x
⇒ 10x = 14 ⇒ x= 7/5
Substituting the value of x in (i), we get
y² + 49/25 = 25
⇒ y² = 25 - 49/25
⇒ y² = (625 - 49)/25
⇒ y² = 576/25
⇒ y = 24/5
Thus,
AC = 2×AM = 2×y = 2×(24/5) m = 48/5 m = 9.6 m
Distance between Reshma and Mandip is 9.6 m.

 

Let A, B and C represent the positions of Ankur, Syed and David respectively. All three boys at equal distances thus ABC is an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is median of ΔABC and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle a metres then BD = a/2 m.
Applying Pythagoras theorem in ΔABD,
AB² = BD² + AD²
⇒ AD² = AB² - BD²  
⇒ AD² = a² - (a/2)²
⇒ AD² = 3a²/4
⇒ AD = √3a/2
OA = 2/3 AD ⇒ 20 m = 2/3 × √3a/2
⇒ a = 20√3 m
Length of the string is 20√3 m.