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From the Given Figure, Prove That: Ap + Bq + Cr = Bp + Cq + Ar Also Show That: Ap + Bq + Cr = 1/2× Perimeter of δAbc. - 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

From the given figure, prove that:
AP + BQ + CR = BP + CQ + AR

Also show that:
AP + BQ + CR = 1/2× Perimeter of ΔABC.

Solution

Since from B, BQ and BP are the tangents to the circle
Therefore, BQ = BP ………..(i)
Similarly, we can prove that
AP = AR …………..(ii)
and CR = CQ ………(iii)
Adding,
AP + BQ + CR = BP + CQ + AR ………(iv)
Adding AP + BQ + CR to both sides
2(AP + BQ + CR) = AP + PQ + CQ + QB + AR + CR
2(AP + BQ + CR) = AB + BC + CA
Therefore, AP + BQ + CR = 1/2× (AB + BC + CA)
AP + BQ + CR = 1/2× perimeter of triangle ABC

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Solution From the Given Figure, Prove That: Ap + Bq + Cr = Bp + Cq + Ar Also Show That: Ap + Bq + Cr = 1/2× Perimeter of δAbc. 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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