Mathematics
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Answer
The quadrilateral ABCD is shown in the figure below:
Since P lies on the bisector of angle B,
∴ P is equidistant from AB and BC ………..(1)
Also, P lies on the bisector of angle C,
∴ P is equidistant from BC and CD ………..(2)
From (1) and (2) we get,
P is equidistant from AB and CD.
Hence, proved that P is equidistant from AB and CD.
Related Questions
Ruler and compasses only may be used in this question. All construction lines and arcs must be clearly shown, and be of sufficient length and clarity to permit the assessment.
(i) Construct a triangle ABC, in which BC = 6 cm, AB = 9 cm, and ∠ABC = 60°.
(ii) Construct the locus of all points, inside △ABC, which are equidistant from B and C.
(iii) Construct the locus of the vertices of the triangles with BC as base, which are equal in area to △ABC.
(iv) Mark the point Q, in your construction, which would make △QBC equal in area to △ABC, and isosceles.
(v) Measure and record the length of CQ.
Construct an isosceles triangle ABC such that AB = 6 cm, BC = AC = 4 cm. Bisect ∠C internally and mark a point P on this bisector such that CP = 5 cm. Find the points Q and R which are 5 cm from P and also 5 cm from the line AB.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
(i) Complete the rectangle ABCD such that :
(a) P is equidistant from AB and BC.
(b) P is equidistant from C and D.
(ii) Measure and record the length of AB.
Use ruler and compasses only for the following question. All construction lines and arcs must be clearly shown.
(i) Construct a △ABC in which BC = 6.5 cm, ∠ABC = 60° and AB = 5 cm.
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