“A diagonal of a parallelogram divides it into two congruent triangles.” Prove it.

Given: ABCD is a parallelogram. AC is a diagonal of parallelogram ABCD which divides it into two triangles, namely, ∆ABC and ∆CDA.

To Prove: ∆ABC ≅ ∆CDA

Proof: BC || DA

| Opposite sides of a parallelogram are parallel

and AC is a transversal

∴ ∠BCA = ∠DAC ...(1)

| Pair of alternate interior angles

Also, AB || DC

Opposite sides of a parallelogram are parallel

and AC is a transversal

∴ ∠BAC = ∠DCA ...(2)

| Pair of alternate interior angles

AC = CA ...(3) | Common

In view of (1), (2) and (3),

∆ABC ≅ ∆CDA

| ASA congruence criterion

1703 Views

Given ∆ABC, lines are drawn through A, B and C parallel respectively to the sides

BC, CA and AB forming ∆PQR. Show that BC =

Given: ∆ABC, lines are drawn through A, B and C parallel respectively to the sides BC, CA and AB forming ∆PQR.

To Prove: BC=

Proof: ∵ AQ || CB and AC || QB

∴ AQBC is a parallelogram

∴ BC = QA ...(1)

| Opposite sides of a ||gm

∵ AR || BC and AB || RC

∴ ARCB is a parallelogram

∴ BC = AR ...(2)

| Opposite sides of a ||gm

From (1) and (2),

From (1) and (3), BC = QR.

1475 Views

In a parallelogram, show that the angle bisectors of two adjacent angles intersect at right angles.

ABCD is a parallelogram. The angle bisectors AE and BE of adjacent angles A and B meet at E.

To Prove: ∠AEB = 90°

Proof: ∵ AD || BC

| Opposite sides of ||gm and transversal AB intersects them

∴ ∠DAB + ∠CBA = 180°

| ∵ Sum of consecutive interior angles on the same side of a transversal is 180°

⇒ 2∠EAB + 2∠EBA = 180°

| ∵ AE and BE are the bisectors of ∠DAB and ∠CBA respectively.

⇒ ∠EAB + ∠EBA = 90° ...(1)

In ∆EAB,

∠EAB + ∠EBA + ∠AEB = 180°

| ∵ The sum of the three angles of a triangle is 180°

⇒ 90° + ∠AEB = 180° | From (1)

⇒ ∠AEB = 90°.

2967 Views

ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively. Show that AX || CY.

Given: ABCD is a parallelogram and line segments AX, CY bisect the angles A and C respectively.

To Prove: AX || CY.

Proof: ∵ ABCD is a parallelogram.

∴ ∠A = ∠C | Opposite ∠s

| ∵ Halves of equals are equal

⇒ ∠1 = ∠2 ....(1)

| ∵ AX is the bisector of ∠A and CY is the bisector of ∠C

Now, AB || DC and CY intersects them

∴ ∠2 = ∠3 ...(2)

| Alternate interior ∠s

From (1) and (2), we get

∠1 = ∠3

But these form a pair of equal corresponding angles

∴ AX || CY.

3477 Views

Advertisement

- Diagonals of a rectangle are perpendicular to each other
- Diagonals of a rhombus are perpendicular to each other
- A kite is a parallelogram whose adjacent sides are equal
- A kite is a parallelogram whose adjacent sides are equal

B.

Diagonals of a rhombus are perpendicular to each other 96 Views

Advertisement

Advertisement