In the given figure, ABCD is a square, EF || BD and R is the mid-point of EF. Prove that:

(i) BE = DF

(ii) AR bisects ∠BAD

(iii) If AR is produced, it will pass through C.

(i) Given,

EF || BD

DC is the transversal.

⇒ ∠BDC = ∠EFC

BC is the transversal.

⇒ ∠DBC = ∠FEC

Given,

ABCD is a square.

⇒ ∠A = ∠B = ∠C = ∠D = 90°

In △BDC,

BC = DC [Sides of a square are equal]

⇒ ∠DBC = ∠CDB = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠DBC + ∠BCD + ∠CDB = 180°

⇒ x + 90° + x = 180°

⇒ 2x = 180° - 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

⇒ ∠DBC = ∠CDB = 45°

∴ ∠DBC = ∠CDB = ∠FEC = ∠EFC = 45°.

⇒ △BDC and △EFC are isosceles triangles.

⇒ EC = CF (As, ∠FEC = ∠EFC)

From figure,

⇒ BE = BC - EC …..(1)

⇒ DF = DC - CF

As, BC = DC and EC = CF

⇒ DF = BC - EC …..(2)

From (1) and (2), we get :

⇒ BE = DF.

Hence, proved that BE = DF.

(ii) In △ABE and △ADF,

⇒ ∠ABE = ∠ADF (Both equal to 90°)

⇒ AB = AD (Sides of a square)

⇒ BE = DF (Proved above)

∴ △ABE ≅ △ADF (By S.A.S axiom)

⇒ ∠BAE = ∠DAF (Corresponding parts of congruent triangles are equal)

⇒ AE = AF (Corresponding parts of congruent triangles are equal)

∴ △AEF is an isosceles triangle.

Given,

R is the mid-point of EF.

We know that,

In an isosceles triangle, the median to the base is also the angle bisector of the vertex angle.

∴ AR bisects ∠EAF.

⇒ ∠EAR = ∠FAR

From figure,

⇒ ∠BAR = ∠BAE + ∠EAR ….(1)

⇒ ∠DAR = ∠DAF + ∠FAR

Since,

∠EAR = ∠FAR and ∠BAE = ∠DAF.

⇒ ∠DAR = ∠BAE + ∠EAR …..(2)

From eq.(1) and (2), we have:

⇒ ∠BAR = ∠DAR.

∴ AR bisects ∠BAD.

Hence, proved that AR bisects ∠BAD.

(iii) Since, AR bisects ∠BAD

AR lies on diagonal AC, as diagonals of a square bisect the vertex angle.

∴ AR if produced, must pass through the vertex C.

Hence, proved that, if AR is produced, it will pass through C.