Question

quadrilateral.
2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
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Answer 1

Given : A parallelogram ABCD , in which AC = BD

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABD

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]⇒ △ABC ≅ △BAD [ by SSS congruence axiom]

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]⇒ △ABC ≅ △BAD [ by SSS congruence axiom]⇒ ∠ABC = △BAD [c.p.c.t.]

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]⇒ △ABC ≅ △BAD [ by SSS congruence axiom]⇒ ∠ABC = △BAD [c.p.c.t.]Also, ∠ABC + ∠BAD = 180° [co - interior angles]

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]⇒ △ABC ≅ △BAD [ by SSS congruence axiom]⇒ ∠ABC = △BAD [c.p.c.t.]Also, ∠ABC + ∠BAD = 180° [co - interior angles]⇒ ∠ABC + ∠ABC = 180° [∵ ∠ABC = ∠BAD]

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]⇒ △ABC ≅ △BAD [ by SSS congruence axiom]⇒ ∠ABC = △BAD [c.p.c.t.]Also, ∠ABC + ∠BAD = 180° [co - interior angles]⇒ ∠ABC + ∠ABC = 180° [∵ ∠ABC = ∠BAD]⇒ 2∠ABC = 180°

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]⇒ △ABC ≅ △BAD [ by SSS congruence axiom]⇒ ∠ABC = △BAD [c.p.c.t.]Also, ∠ABC + ∠BAD = 180° [co - interior angles]⇒ ∠ABC + ∠ABC = 180° [∵ ∠ABC = ∠BAD]⇒ 2∠ABC = 180° ⇒ ∠ABC = 1 /2 × 180° = 90°

Given : A parallelogram ABCD , in which AC = BD TO Prove : ABCD is a rectangle .Proof : In △ABC and △ABDAB = AB [common]AC = BD [given]BC = AD [opp . sides of a | | gm]⇒ △ABC ≅ △BAD [ by SSS congruence axiom]⇒ ∠ABC = △BAD [c.p.c.t.]Also, ∠ABC + ∠BAD = 180° [co - interior angles]⇒ ∠ABC + ∠ABC = 180° [∵ ∠ABC = ∠BAD]⇒ 2∠ABC = 180° ⇒ ∠ABC = 1 /2 × 180° = 90° Hence, parallelogram ABCD is a rectangle.

Answer 2

Step-by-step explanation:

Gven: In parallelogram ABCD, AC=BD

To prove : Parallelogram ABCD is rectangle.

Proof : in △ACB and △BDA

AC=BD ∣ Given

AB=BA ∣ Common

BC=AD ∣ Opposite sides of the parallelogram ABCD

△ACB ≅△BDA∣SSS Rule

∴∠ABC=∠BAD...(1) CPCT

Again AD ∥ ∣ Opposite sides of parallelogram ABCD

AD ∥BC and the traversal AB intersects them.

∴∠BAD+∠ABC=180∘

...(2) _ Sum of consecutive interior angles on the same side of the transversal is

180∘

From (1) and (2) ,

∠BAD=∠ABC=90∘

∴∠A=90∘

and ∠C=90∘

Parallelogram ABCD is a rectangle.

and ∠C=90∘

Parallelogram ABCD is a rectangle.