Given: Chord AB of the outer circle cuts the inner circle at C and D.

To prove: AB − CD = 2AC
Construction: Draw OM ⊥ AB
Proof: Since, OM ⊥ AB (By construction)
Also, OM ⊥ CD
Therefore, AM = MB ... (1) (Perpendicular drawn from centre to chord bisect it)
CM = MD ... (2)
Now, AB – CD = 2AM–2CM
= 2(AM–CM) from (1) and (2)
AB – CD = 2AC

The distance between two chords FE
= OE + OF
OE = \(\sqrt { { 10 }^{ 2 }-{ 6 }^{ 2 } } =\sqrt { 100-36 } =\sqrt { 64 } =8cm\)
OF = \(\sqrt { { 10 }^{ 2 }-{ 8 }^{ 2 } } \)
= \(\sqrt { 100-64 } \)
= \(\sqrt { 36 } =6cm\)
\(\therefore \) FE = 8 cm + 6 cm = 14 cm
\(\therefore\) Distance between the chords is 14 cm

BD = \(\sqrt { { 15 }^{ 2 }+{ 12 }^{ 2 } } \)
= \(\sqrt { 225-144 } \)
= \(\sqrt { 81 } \)
= 9 cm
\(\therefore\) length of the chord AB = 9 + 9 = 18

\(\triangle \)OAC is an isosceles triangle with one angle 90°
\(\therefore \ \angle OAC+\angle OCA={ 180 }^{ 0 }-{ 90 }^{ 0 }\)
\(2\angle OAC={ 90 }^{ 0 }\)   \(\left( \therefore \ \angle OAC=\angle OCACA \right) \)
\(\angle OAC={ 45 }^{ 0 }\)
\(\therefore \angle OAC={ 45 }^{ 0 }\)
Length of the chord =\(\sqrt { \left( 4\sqrt { 2 } \right) ^{ 2 }+\left( 4\sqrt { 2 } \right) ^{ 2 } } \)
=\(\sqrt { 16\times 2+16\times 2 } \)
=\(\sqrt { 64 } =8cm\)

In an equilateral \(\triangle \) on side = 6 cm,

Construction :
Step 1: Draw ΔABC with equal sides of 6 cm length.
Step 2: Draw perpendicular bisectors of any two sides (BC and AC) to find the midpoints of BC and AC.
Step 3: Draw medians BD and CE. Let them meet at G.
Step 4: Draw angle bisectors of any two sides (ㄥB and ㄥC): Let them meet at I.
Step 5: G and I are concurrent point and G is the centroid, I is the incentre.
Step 6: In an equilateral triangle centroid and incentre lie on the same point.

In \(\triangle \)ABC, AB = 6 cm, ∠B= 110°, AC = 9 cm

Construction :
Step 1: Draw \(\triangle \)ABC with AB = 6 cm, ∠B =110°, AC = 9 cm
Step 2: Draw perpendicular bisectors of any two sides (BC and AB) to find the midpoints of BC and AB.
Step 3: Construct medians AD and CE. Let them meet at G.
Step 4: G is the centroid of the given \(\triangle \)ABC.

In \(\triangle \)LMN
LM = 7.5 cm,
MN = 5 cm,
LN = 8 cm

Construction :
Step 1: Draw \(\triangle \)LMN with LNM = 8 cm, MN = 5 cm, LM = 7.5 cm
Step 2: Construct perpendicular bisectors for any two sides (LN and MN) to find the midpoints of LM and MN.
Step 3: Draw the medians LD, ME. Let them meet at G.
Step 4: G is the centroid of the triangle LMN.