Question

10. In a parallelogram ABCD, the angle bisector of
angleA bisects BC. Will angle bisector of angleB also
bisects AD? Give reason.
please answer it fast with complete solve​

Answer 1

Answer:

ANSWER

ANSWER (i) Let AD = xAB = 2AD = 2x

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = x

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = xAlso, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = xAlso, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)Also, ∠6 = ∠3 (alternate angles)

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = xAlso, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)Also, ∠6 = ∠3 (alternate angles)∵ ∠6 = ∠4 and ∠6 = ∠3⇒∠3 = ∠4

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = xAlso, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)Also, ∠6 = ∠3 (alternate angles)∵ ∠6 = ∠4 and ∠6 = ∠3⇒∠3 = ∠4Hence, BP bisects ∠B.

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = xAlso, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)Also, ∠6 = ∠3 (alternate angles)∵ ∠6 = ∠4 and ∠6 = ∠3⇒∠3 = ∠4Hence, BP bisects ∠B. (ii) To prove ∠APB = 90°∵ Opposite angles are supplementary..

ANSWER (i) Let AD = xAB = 2AD = 2xAlso AP is the bisector ∠A∴∠1 = ∠2Now, ∠2 = ∠5 (alternate angles)∴∠1 = ∠5Now AD = DP = x [∵ Sides opposite to equal angles are also equal]∵ AB = CD (opposite sides of parallelogram are equal)∴ CD = 2x⇒ DP + PC = 2x⇒ x + PC = 2x⇒ PC = xAlso, BC = x In ΔBPC,∠6 = ∠4 (Angles opposite to equal sides are equal)Also, ∠6 = ∠3 (alternate angles)∵ ∠6 = ∠4 and ∠6 = ∠3⇒∠3 = ∠4Hence, BP bisects ∠B. (ii) To prove ∠APB = 90°∵ Opposite angles are supplementary.. Angle sum property,

Answer 2

Answer:

In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Prove that AD = 2AB. In a parallelogram ABCD, the bisector of ∠A also bisects BC at X. Prove that AD = 2AB