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In the Given Figure, Ac = Ae Show That: (I) Cp = Ep (Ii) Bp = Dp - ICSE Class 10 - Mathematics

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ConceptTangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments

Question

In the given figure, AC = AE Show that:
(i) CP = EP
(ii) BP = DP

 

Solution

In ΔADC and ΔABE ,
`∠`ACD = `∠`AEB (angles in the same segment)
AC = AE (Given)
`∠`A = `∠`A(common)
∴ Δ ADC ≅  Δ ABE (ASA postulate)
 ⇒ AB = AD
But AC = AE
∴ AC – AB = AE – AD
⇒  BC = DE
In Δ BPC and ΔDPE
`∠`C =  `∠`E (angles in the same segment)
BC = DE
`∠`CBP = `∠`CDE (angles in the same segment)
∴ ΔBPC ≅  ΔDPE (ASA Postulate)
⇒ BP = DP and CP = PE (cpct)

  Is there an error in this question or solution?

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Solution In the Given Figure, Ac = Ae Show That: (I) Cp = Ep (Ii) Bp = Dp Concept: Tangent Properties - If a Line Touches a Circle and from the Point of Contact, a Chord is Drawn, the Angles Between the Tangent and the Chord Are Respectively Equal to the Angles in the Corresponding Alternate Segments.
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