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1.

Theorem: The length of two tangents drawn from an external point to a circle are equal.

Answer» In `Delta PAD` and `Delta QAD`,
seg `PA ~=`seg QA …(Radii of the same circle )
seg `AD ~=` seg AD …...(Common side )
`/_ APD = /_AQD = 90^(@)` …(Tangent theorem )
`:. Delta PAD ~=Delta QAD ` ....(Hypotenuse side theorem )
`:.` seg `DP ~=` seg DQ ....(c.s.c.t. )
Discussion
2.

If two chords of a congruent circle are equal; then their corresponding arcs.

Answer» In `Delta AOB ` and `Delta COD`,
seg OA `~=` seg OC
seg `OB ~= `seg OD ….(Radii of the same circle )
seg AB `~=` seg CD …...(Given )
`:. Delta AOB ~= Delta COD ` …..(SSS test of congruence )
`:. /_AOB ~= /_COD ` ....(c.a.c.t.) ......(1)
m ( arc AMB ) `= /_ AOB ` .....(Definiton of measure of minor arc ) ........(2)
m (arc CND ) =`/_COD ` .....(3)
`:.` m (arc AMB ) m ( arc CND ) .....[From (1) , (2) and (3) ]
`:.` ar AMB `~=` arc CND .
Discussion
3.

In the figure, O is the centre of the circle. Seg AC is the diameter `/_ ABC` is inscribed in arc ABC and intercepts arc AMC then prove `/_ ABC = 90^(@)`

Answer» Proof `:` m (arc AMC ) `= 180^(@)` …(Measure of a semicircle is `180^(@)`) …(1)
`/_ABC = (1)/(2) `m (arc AMC ) …(Inscribed angle theorem )
`:. /_ ABC = (1)/(2) xx 180^(@)` ….[From (1) ]
`:. /_ABC = 90^(@)`
Discussion
4.

Theorem of internal division of chords. Suppose two chords of a circle intersect each other in the interior of the circle, then the product of the lengths of the two segments of one chord is equal to the product of the lengths of the two segments of the other chord. Given `:` (1) A circle with centre O . (2) chords PR and QS intersect at point E inside the circle. To prove `:` PE `xx` ER = QE xx ES ` Construction `:` Draw seg PQ and seg RS

Answer» Proof `:` `/_PQS ~= /_SRP ` ....(Angles inscribed in the same arc ar congruent )
i.e., `/_PQE ~= /_SRE ` ….(P-E-R and Q-E-S ) ….(1)
In `Delta PQE` and `Delta SRE`,
`/_PQE ~= /_SRE` ….[From (1) ]
`/_ PEQ ~= /_SER` .....(Vertically opposite angles )
`:. Delta PQE ~ Delta SRE ` .....(AA test of similarity )
` :. ( PE )/( SE) = (QE)/(RE)` ......(Corresponding sides of similar triangles are in proportion )
`:. PE xx ER = QE xx SE `.
Discussion
5.

Corollaries of inscribed angle theorem `:` Angle inscribed in the same arc arc contruent Given `:` (1) A circle with centre O (2) `/_ABD` and `/_ACD ` are inscribed in arc ABC and intercepts arc APD. To prove `:` `/_ABD ~= /_ACD `

Answer» Proof `:` `/_ABD = (1)/(2)` m (arc APD ) ...(1)
`/_ACD = (1)/(2) ` m (arc APD ) Inscribed angle theorem ) …(2)
From (1) and (2), we get
`/_ABD ~= /_ACD `
Discussion
6.

Prove that , "If a line parallel to a side of a triangle intersects the remaining sides in two distinct points then the line divides the sides in the same proportion".

Answer» Proof `:` `Delta XAB ` and `Delta BAY ` have a common vertex B and their bases XA and AY lie on the same line XY.
`:.` they have equal heights.
`:. (A (Delta XAB))/(A ( Delta BAY )) = ( XA )/( AY ) ` …[Triangles of equal heights ] ….(1)
`Delta XAB ` and `Delta ABZ ` have a common verte A and their bases XB and BZ lie on the same line XZ.
`:.` they have equal heights.
`:. (A( Delta XAB))/( A ( Delta ABZ)) = ( XB)/( BZ)` ....[Triangles of equal heights] ....(2)
`:. Delta BAY ` and `Delta ABZ ` lie between the same two parallel lines AB and YZ.
`:. ` they have equal heights, also have same base AB.
`:. A ( Delta BAY ) = A ( Delta ABZ ) `.... [Triangles of equal bases and heights ] ....(3)
`:.` from (1) , (2) and (3) , we get
`(A(DeltaXAB))/(A(DeltaBAY)) = ( A(Delta XAB))/(A(DeltaABZ))` ....(4)
`:.` from (1) ,(2) and (4) , we get
`( XA)/( AY) = (XB)/( BZ)`
Discussion