Question

In the given figure above, ABCD is a square and EF is || diagonal BD and EM=FM. Prove that:

1) BE=DF

2) AM bisects angle BAD.​

Answer 1

Answer:

Solution:-

(1) Since diagonal of a square bisects the vertex and BD is the diagonal of square ABCD.

∴ ∠ CBD = ∠ CDB = 90/2 = 45°

Given : EF || BD

⇒ ∠ CEF = ∠ CBD = 45° and ∠ CEF = ∠ CDB = 45° (Corresponding angles)

⇒ CEF = CFE

⇒ CE = CF (Sides opposite of equal angles are equal) .....(1)

Now, BC = CD (Sides of square) .....(2)

Subtracting (1) from (2), we get

⇒ BC CE = CD CF

⇒ BE = DF or DF = BE (First condition proved)

(2) Δ ABE ≡ ADF (By SAS congruency criterion)

⇒ ∠ BAE = ∠ DAF .....(3)

AE = AF

And, Δ AEM ≡ Δ AFM (By SSS congruency criterion)

⇒ ∠ EAM = ∠ FAM ....(4)

Now adding (3) and (4), we get

⇒ BAE + EAM = DAF + FAM

⇒ BAM = DAM

i.e. AM bisects ∠ BAD

Proved.

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Answer 2

$\huge\mathfrak\purple{Bonjour!!}$

$\LARGE\underline{\underline{\sf\red{G}\blue{i}\green{v}\orange{e}\red{n}:}}$

⛄ ABCD is a square.

⛄ EF is || BD.

⛄ EM = FM.

$\LARGE\underline{\underline{\sf\red{T}\blue{o}\green\:{P}\orange{r}\red{o}\blue{v}\green{e}:}}$

❄ BE = DF.

❄ AM bisects angle BAD.

$\LARGE\underline{\underline{\sf\red{S}\blue{o}\green{l}\orange{u}\red{t}\blue{i}\green{o}\orange{n}:}}$

⏩ 1) Since ABCD is a square,

BC=CD -----> (i) [All sides of a square are equal in measure]

The diagonal BD bisects angle B and angle D. Therefore,

angle B = angle D = 90°

=> angle 1 = angle 2 = 45°

=> angle 3 = angle 2 and angle 4 = angle 1 [Corresponding angles]

=> angle 3 = angle 4

Therefore,

∆CEF is an isosceles triangle.

=> CE = CF -----> (ii) [Sides opposite to equal angles are equal]

Now, on subtracting (ii) from (i), we get,

BC - CE = CD - CF

=> BE = DF.

⏩ 2) Now, consider ∆ADF and ∆ABE. Here,

AD = AB [Sides of a square ]

angle D= angle B [90°]

DF = BE [Proved above]

Therefore, by SAS rule, ∆ADF is congruent to ∆ABE.

=> AF = AE.

=> x° = y° ---------> (iii).

Now, in ∆AFM and ∆AEM,

AF = AE [Proved above]

AM = AM [Common]

FM = ME [Given]

Therefore, by SSS rule, ∆AFM is congruent to ∆AEM.

=> t° = s° --------- (iv)

Now, on adding (iii) and (iv), we get,

x° + t° = y° + s°

i.e, angle DAM = angle BAM

and

=> AM bisects angle BAD.

[Do have a glimpse at the attachment to view the figure!]

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