Question

10) In a quadrilateral ABCD, diagonals AC and BD are equal perpendicular to each other. What type of
quadrilateral is ABCD
(a) Square (b) Parallelogram (c) Rhombus
(d) Trapezium​

Answer 1

Step-by-step explanation:

⭐Question;-

In a quadrilateral ABCD, diagonals AC and BD are equal perpendicular to each other. What type of

quadrilateral is ABCD

⭐Answer:-

Given:-

AE : EC = BE : ED

Draw EF || AB

In ΔABD, EF || AB

Using Basic Proportionality theorem,

$\frac{DF}{FA} = \frac{DE}{EB}$

$but \: given \: \frac{de}{EB} = \frac{CE}{CA}$

∴$\frac{DF}{FA} = \: \frac{CE}{CA}$

Thus, in ΔDCA, E and F are points on CA and DA respectively such that

$\frac{DF}{FA} = \frac{CE}{CA}$

Thus, by converse of Basic proportionality theorem, FE || DC.

But, FE || AB.

Hence, AB || DC.

Thus, ABCD is a trapezium.

Answer 2

Answer:-

(a) Square

Proof:-

In given figure ABCD is a quadrilateral having all sides are equal and opposite sides parallel to each other. AC and BD are diagonals.

⇒ In △ABC and △BAD,

⇒ AB = AB [Common line]

⇒ BC = AD [Sides of square are equal.]

⇒ ∠ABC = ∠BAD [All angles of square is 90°]

Therefore, △ABC ≅ △BAD [By SAS property]

In a △ OAD and △OCB,

⇒ AD = CB [sides of a square]

⇒ ∠OAD = ∠OCB [Alternate angle]

⇒ ∠ODA = ∠OBC [Alternate angle]

⇒ △OAD ≅ △OCB [By ASA Property]

So, OA = OC ----- ( 1 )

Similarly, OB = OD ------ ( 2 )

From ( 1 ) and ( 2 ) we get that AC and BD bisect each other.

⇒ Now, in △OBA and △ODA,

⇒ OB = OD [From ( 2 )]

⇒ BA = DA

⇒ OA = OA [Common line]

⇒ ∠AOB + ∠AOD ----- ( 3 ) [By CPCT]

⇒ ∠AOB + ∠AOD = 180°

[Linear pair]

⇒ 2∠AOB = 180°

∴ ∠AOB = ∠AOD = 90°

∴ We have proved that diagonals of square are equal and perpendicular.