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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. - Mathematics

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प्रश्न

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.

योग

उत्तर

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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अध्याय 14: Loci (Locus and its Constructions) - Figure Based Questions

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आईसीएसई Mathematics [English] Class 10
अध्याय 14 Loci (Locus and its Constructions)
Figure Based Questions | Q 18

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