ABC is an equilateral triangle. P is a point on AB and Q is a point on AB produced. Prove that, :-
CP < AB
CQ > AB
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Answer:
REF. Image.
Given AB=AC
and AP=AQ
Thus
AB-AP=AC-AQ
[BP=CA ] [from figure ]
now InΔBCP & ΔBCQ
BP = CQ
∠c=∠c [common]
and BC=BC [common]
∴ΔBCP≃ΔBCQ [SAS congruency]
now
[BQ=CP] [corresponding parts of congruent triangles]
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