Line l is the bisector of an angle ∠A and B is any point on l. BP an

1.	Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:(i) ΔAPB ≅ ΔAQB(ii) BP = BQ or B is equidistant from the arms of ∠A.
Answer» Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that: (i) ΔAPB ≅ ΔAQB (ii) BP = BQ or B is equidistant from the arms of ∠A.

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:(i) ΔAPB ≅ ΔAQB(ii) BP = BQ or B is equidistant from the arms of ∠A.

Answer»

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

Discussion

No Comment Found

Related InterviewSolutions

Given, tan(A+B)=√3 and tan(A−B)=1√3If B<A<90∘, then, find the values of A and B in degrees.
Represent -25.6875 in rational form.
Factorise:25x2 – 10x + 1 – 36y2
1.How many pairs of adjacent angles are formed when two lines intersect in a point?
if x=squareroot3 + 1 /2 then value of 4x^3 +2x^2-8x+7 is??
Question 1In the figure, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).
Question 1(ii) Do the following pair of linear equations have no solution? Justify your answer. x = 2y and y = 2x
In a parallelogram ABCD, ∠D = 135°, determine the measures of ∠A and ∠B.
The value of (2x−3y)3+(x+3y)3+3(2x−3y)2(x+3y)+3(2x−3y)(x+3y)2=
Number of real zero(s) of polynomial f(y) = ay2 + by + c, is

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:(i) ΔAPB ≅ ΔAQB(ii) BP = BQ or B is equidistant from the arms of ∠A.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment