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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.
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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