Question

solve question number 10 it's argent plz solve and send me fast and elaborate this question is very important questions it's argent right now

Answer 1

In △AOE and △EOC
OA = OC [radii of same circle]
OE = OE [common]
AE = EC [Tangents drawn from an external point to a circle are equal]
△AOE ≅ △EOC [By Side Side Side Criterion]
∠AEO = ∠CEO [Corresponding parts of congruent triangles are equal ]
∠AEC = ∠AEO + ∠CEO = ∠AEO + ∠AEO = 2∠AEO [1]
Now As CD is a straight line
∠AED + ∠AEC = 180° [linear pair]
2∠AEO = 180 - ∠AED [From]
∠AEO=90-1/2 ∠AED
Now, In △O'ED and △O'EB
O'B = O'D [radii of same circle]
O'E = O'E [common]
EB = ED [Tangents drawn from an external point to a circle are equal]
△O'ED ≅ △O'EB [By Side Side Side Criterion]
∠O'EB = ∠O'ED [Corresponding parts of congruent triangles are equal ]
∠DEB = ∠O'EB + ∠O'ED = ∠O'ED + ∠O'ED = 2∠O'ED [3]
Now as AB is a straight line
∠AED + ∠DEB = 180 [Linear Pair]
2∠O'ED = 180 - ∠AED [From 3]
Hence So AC and BD lies on same line by converse of linear pair.

Answer 2

Here AB and CD are extended then they are intersecting at P.

Length of two tangents drawn from external point to a circle are equal.

AP = CP ----------(1)
BP = DP ----------(2)

Substracting (2) from (1) we get

AP - BP = CP - DP

∴ AB = CD