Question

In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :
(i) AE = AD,
(ii) DE bisects and ∠ADC and
(iii) Angle DEC is a right angle.​

Answer 1

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In parallelogram ABCD, E is the mid-point of side AB and CE bisects angle BCD. Prove that :

(i) AE = AD,

(ii) DE bisects and ∠ADC and

(iii) Angle DEC is a right angle.

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$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

A trapezium ABCD in which AB || CD and AD = BC.

${\blue{\sf\underline{To prove}}}$

(i)∠A = ∠B and ∠C = ∠D,

(ii) ∆ABC ≅ ∆BAD and

(iii) AC = BD.

${\blue{\sf\underline{Construction}}}$

Produce AB tp X. Draw CN || DA such that CN meets AX at N.

${\blue{\sf\underline{Proof}}}$

We have

⠀⠀⠀⠀⠀AD || NC and AN || DC (since AB || DC)⇒ ADCN is a ||gm.

∴ NC = AD = BC ⇒ ∠5 = ∠6.

Now, ∠1 + ∠6 = 180° (co-interior ∠s)

and ∠2 + ∠5 = 180° (linear pair)

∴ ∠1 = ∠2 [using (i)].

Now, ∠3 = ∠6 (opp.∠s of a ||gm)

and ∠4 = ∠5 (alt. interior ∠s)

∴ ∠3 = ∠4 [using (i)]>

Thus,∠A = ∠B and ∠C = ∠D .

Now, in ∆ABC and BAD, we have;

⠀⠀⠀⠀⠀AB = BA (commom)

⠀⠀⠀⠀⠀∠ABC = ∠BAD (prove that ∠A = ∠B)

⠀⠀⠀⠀⠀AD = BC (given)

∴ ∆ABC ≅ ∆BAD (SAS-criterion).

And so, AC = BD (c.p.c.t.).

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Answer 2

Given

A trapezium ABCD in which AB || CD and AD = BC.

To prove

(i)∠A = ∠B and ∠C = ∠D,

(ii) ∆ABC ≅ ∆BAD and

(iii) AC = BD.

Construction

Produce AB tp X. Draw CN || DA such that CN meets AX at N.

Proof

We have

⠀⠀⠀⠀⠀AD || NC and AN || DC (since AB || DC)⇒ ADCN is a ||gm.

∴ NC = AD = BC ⇒ ∠5 = ∠6.

Now, ∠1 + ∠6 = 180° (co-interior ∠s)

and ∠2 + ∠5 = 180° (linear pair)

∴ ∠1 = ∠2 [using (i)].

Now, ∠3 = ∠6 (opp.∠s of a ||gm)

and ∠4 = ∠5 (alt. interior ∠s)

∴ ∠3 = ∠4 [using (i)]>

Thus,∠A = ∠B and ∠C = ∠D .

Now, in ∆ABC and BAD, we have;

⠀⠀⠀⠀⠀AB = BA (commom)

⠀⠀⠀⠀⠀∠ABC = ∠BAD (prove that ∠A = ∠B)

⠀⠀⠀⠀AD = BC (given)

∴ ∆ABC ≅ ∆BAD (SAS-criterion).

And so, AC = BD (c.p.c.t.).