Advertisements
Advertisements
Question
In Δ PQR, bisectors of ∠ PQR and ∠ PRQ meet at I. Prove that I is equidistant from the three sides of the triangle , and PI bisects ∠ QPR .
Solution
Since I lies on bisector of ∠ R, I is equidistant from PR and QR.
Again I lies on the bisector of ∠ Q , I is equidistant from PQ and QR.
Hence , I is equi distant from alI sides of the triangle.
Therefore, I lies on the bisector of ∠ P i.e ∠ QPR .
APPEARS IN
RELATED QUESTIONS
On a graph paper, draw the lines x = 3 and y = –5. Now, on the same graph paper, draw the locus of the point which is equidistant from the given lines.
On a graph paper, draw the line x = 6. Now, on the same graph paper, draw the locus of the point which moves in such a way that its distantce from the given line is always equal to 3 units
Describe the locus of a point P, so that:
AB2 = AP2 + BP2,
where A and B are two fixed points.
Construct a triangle BCP given BC = 5 cm, BP = 4 cm and ∠PBC = 45°.
- Complete the rectangle ABCD such that:
- P is equidistant from AB and BC.
- P is equidistant from C and D.
- Measure and record the length of AB.
Two straight roads AB and CD cross each other at Pat an angle of 75° . X is a stone on the road AB, 800m from P towards B. BY taking an appropriate scale draw a figure to locate the position of a pole, which is equidistant from P and X, and is also equidistant from the roads.
In the given figure ABC is a triangle. CP bisects angle ACB and MN is perpendicular bisector of BC. MN cuts CP at Q. Prove Q is equidistant from B and C, and also that Q is equidistant from BC and AC.
Without using set squares or protractor construct a triangle ABC in which AB = 4 cm, BC = 5 cm and ∠ABC = 120°.
(i) Locate the point P such that ∠BAp = 90° and BP = CP.
(ii) Measure the length of BP.
State and draw the locus of a point equidistant from two given parallel lines.
Construct a Δ ABC, with AB = 6 cm, AC = BC = 9 cm; find a point 4 cm from A and equidistant from B and C.
Using a ruler and compass only:
(i) Construct a triangle ABC with BC = 6 cm, ∠ABC = 120° and AB = 3.5 cm.
(ii) In the above figure, draw a circle with BC as diameter. Find a point 'P' on the circumference of the circle which is equidistant from Ab and BC.
Measure ∠BCP.