ABCD is a rhombus EABF is a straight line such that EA=AB=BF Prove that ED and FC when produced meet - Maths - Quadrilaterals

Question

ABCD is a rhombus EABF is a straight line such that EA=AB=BF Prove
that ED and FC when produced meet at right angle

Vimala · Accepted Answer

Dear student, Consider X as O according to the figure in the question provided Let us observe the following figure

Consider the triangles, $Δ E A D a n d Δ A B C$ Since the sides EA = AB (Given), $∠ E A D = ∠ A B C$ (corresponding angles), and AD = BC (sides of the rhombus), by Side-Angle-Side congruency, the triangles $Δ E A D a n d Δ A B C$ are congruent Therefore, the corresponding parts of corresponding triangles are also equal Thus, $∠ A E D = ∠ B A C$ Consider the triangles, $Δ C B F a n d Δ D A B$ Since the sides BF = AB (Given), $∠ C B F = ∠ D A B$ (corresponding angles), and AD = BC (sides of the rhombus), by Side-Angle-Side congruency, the triangles $Δ C B F a n d Δ D A B$ are congruent Therefore, the corresponding parts of corresponding triangles are also equal Thus, $∠ C F B = ∠ D B A$ In $Δ A B O$ , $∠ A B O + ∠ B A O + ∠ A O B = 180^{\circ} \Rightarrow ∠ A B O + ∠ B A O = 90^{\circ} [I n R hom b u s, d i a g o n a l s b i \sec t a t 90^{\circ}] \Rightarrow ∠ C F B + ∠ A E D = 90^{\circ} [∠ D B A = ∠ A B O = ∠ C F B, ∠ A E D = ∠ B A O = ∠ B A C] \Rightarrow ∠ X F E + ∠ X E F = 90^{\circ}$ We have, $∠ E X F + ∠ X F E + ∠ X E F = 180^{\circ}$ Thus, $∠ E X F = 90^{\circ}$

ABCD is a rhombus.EABF is a straight line such that EA=AB=BF.Prove that ED and FC when produced meet at right angle