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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

In the adjoin...

Question

In the adjoining figure P and Q are two points on equal sides AB and AC of an isosceles triangle ABC such that AP $=$ AQ , prove that BQ $=$ CP

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Solution

REF. Image.
Given AB=AC
and AP=AQ
Thus
AB-AP=AC-AQ
[BP=CA ] [from figure ]
now In $Δ B C P$ & $Δ B C Q$
BP = CQ
$∠ c = ∠ c$ [common]
and BC=BC [common]
$∴ Δ B C P ≃ Δ B C Q$ [SAS congruency]
now
[BQ=CP] [corresponding parts of congruent triangles]

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In the adjoining figure P and Q are two points on equal sides AB and AC of an isosceles triangle ABC such that AP = AQ , prove that BQ = CP

REF. Image.Given AB=ACand AP=AQThusAB-AP=AC-AQ[BP=CA ] [from figure ]now InΔBCP & ΔBCQBP = CQ∠c=∠c [common]and BC=BC [common]∴ΔBCP≃ΔBCQ [SAS congruency]now[BQ=CP] [corresponding parts of congruent triangles]

In the adjoining figure P and Q are two points on equal sides AB and AC of an isosceles triangle ABC such that AP $=$ AQ , prove that BQ $=$ CP

REF. Image.
Given AB=AC
and AP=AQ
Thus
AB-AP=AC-AQ
[BP=CA ] [from figure ]
now In $Δ B C P$ & $Δ B C Q$
BP = CQ
$∠ c = ∠ c$ [common]
and BC=BC [common]
$∴ Δ B C P ≃ Δ B C Q$ [SAS congruency]
now
[BQ=CP] [corresponding parts of congruent triangles]