Mathematics
In the given figure, P is mid-point of side AD of rectangle ABCD; then :
∠PBC = ∠PBA
∠PBC = ∠PCB
∠BPA = ∠BPC
∠PBC = ∠BPA
Answer
In △ PCD and △ PBA,
⇒ ∠D = ∠A (Both equal to 90°)
⇒ AP = PD (P is mid-point of AD)
⇒ DC = AB (Opposite sides of rectangle are equal)
∴ △ PCD ≅ △ PBA (By S.A.S. axiom)
We know that,
Corresponding sides of congruent triangle are equal.
∴ ∠PCD = ∠PBA
⇒ 90° - ∠PCD = 90° - ∠PBA
⇒ ∠PCB = ∠PBC.
Hence, Option 2 is the correct option.
Related Questions
From the given figure, if ∠A = ∠C, we get :
x = 8, y = 16
x = -8, y = 16
x = 16, y = -8
x = 16, y = 8
ABCD is a rectangle. X and Y are points on sides AD and BC respectively such that AX = BY, then :
AY ≠ BX
△ ABX ≅ △ BYA
△ ABX ≅ △ AYB
△ ABX ≅ △ BAY
In the given figure, AB = AC. Prove that :
(i) DP = DQ
(ii) AP = AQ
(iii) AD bisects angle A
In triangle ABC, AB = AC; BE ⊥ AC and CF ⊥ AB. Prove that :
(i) BE = CF
(ii) AF = AE
