Advertisements
Advertisements
Question
In parallelogram ABCD, the angle bisector of ∠A bisects BC. Will angle bisector of B also bisect AD? Give reason.
Advertisements
Solution
Given, ABCD is a parallelogram, bisector of ∠A, bisects BC at F, i.e. ∠1 = ∠2, CF = FB.
Draw FE || BA.
ABFE is a parallelogram by construction ...[∵ FE || BA]
⇒ ∠1 = ∠6 ...[Alternate angle]
But ∠1 = ∠2 ...[Given]
∴ ∠2 = ∠6
AB = FB [Opposite sides to equal angles are equal] ...(i)
∴ ABFE is a rhombus.
Now, In ΔABO and ΔBOF,
AB = BF ...[From equation (i)]
BO = BO ...[Common]
AO = FO ...[Diagonals of rhombus bisect each other]
∴ ΔABO ≅ ΔBOF ...[By SSS]
∠3 = ∠4 ...[By CPCT]
Now, BF = `1/2` BC ...[Given]
⇒ BF = `1/2` AD ...[BC = AD]
⇒ AE = `1/2` AD ...[BF = AE]
∴ E is the midpoint of AD.
APPEARS IN
RELATED QUESTIONS
The following figure GUNS is a parallelogram. Find x and y. (Lengths are in cm)
Find the measure of ∠P and ∠S, if `bar(SP) || bar(RQ)` in the following figure. (If you find m∠R, is there more than one method to find m∠P?).
Name the quadrilaterals whose diagonals bisect each other
In the given figure, `square`PQRS and `square`ABCR are two parallelograms. If ∠P = 110° then find the measures of all angles of `square`ABCR.
In the given figure, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that `square`PQRS is a parallelogram.
ABCD is a parallelogram. What kind of quadrilateral is it if : AC = BD and AC is perpendicular to BD?
In parallelogram ABCD, E is the mid-point of AD and F is the mid-point of BC. Prove that BFDE is a parallelogram.
Iron rods a, b, c, d, e, and f are making a design in a bridge as shown in the figure. If a || b, c || d, e || f, find the marked angles between b and c
Iron rods a, b, c, d, e, and f are making a design in a bridge as shown in the figure. If a || b, c || d, e || f, find the marked angles between d and f
Construct a parallelogram ABCD in which AB = 4 cm, BC = 5 cm and ∠B = 60°.
