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Question
ΔPBC and ΔQBC are two isosceles triangles on the same base BC but on the opposite sides of line BC. Show that PQ bisects BC at right angles.
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Solution
Given: Two ΔSPBC and QBC on the same base BC but in the opposite sides of BC such that PB = PC and QB = QC.
To prove: PQ bisects BC and is ⊥ to BC.
Proof: Since, the locus of points equidistant from two given points is the perpendicular bisector of the segment joining them. Therefore, ΔPBC is isoceles
⇒ P lies on the perpendicular bisector of BC
ΔQBC is isoceles ⇒ QB = QC
⇒ Q lies on the perpendicular bisectors of BC
∴ PQ is the perpendicular bisectors of BC
Hence, PQ bisects BC at right angles.
Hence proved.
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