Question

In the adjacent figure in which AD is perpendicular to BC and AB = p, AC = q, BD = r, DC = s then show that p^2-q^2=r^2-s^2

Answer 1

$\huge\bold{\gray{\sf{Answer:}}}$

$\bold{Explanation:}$

Quadrilateral:

The

closed figure formed by joining four non collinear points in an order is called

a quadrilateral.

·Trapezium:

A quadrilateral in

which one pair of opposite sides are parallel is called a trapezium.

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Given, ABCD is a trapezium in which AB||CD &

AD=BC

To Show:

(i) ∠A = ∠B

(ii) ∠C = ∠D

(iii) ΔABC ≅ ΔBAD

(iv) diagonal AC = diagonal BD 

Construction: Draw a line through C parallel to DA

intersecting AB produced at E.

Proof:

i)

AB||CD(given)

AD||EC (by construction)

So ,ADCE is a parallelogram

CE = AD

(Opposite sides of a parallelogram)

AD = BC (Given)

We know that ,

∠A+∠E= 180°

[interior

angles on the same side of the transversal AE]

∠E= 180° - ∠A

Also, BC = CE

∠E = ∠CBE= 180° -∠A

∠ABC= 180° - ∠CBE

[ABE  is a straight line]

∠ABC= 180° - (180°-∠A)

∠ABC= 180° - 180°+∠A

∠B= ∠A………(i)

 

(ii) ∠A + ∠D = ∠B + ∠C = 180°

 (Angles on

the same side of transversal)

∠A + ∠D = ∠A + ∠C

 (∠A = ∠B) from eq (i)

 ∠D = ∠C

 

(iii) In ΔABC and ΔBAD,

AB = AB (Common)

∠DBA = ∠CBA(from eq (i)

AD = BC (Given)

ΔABC ≅ ΔBAD

 (by SAS congruence rule)

(iv)  Diagonal AC = diagonal BD

 (by CPCT as ΔABC ≅ ΔBAD)

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Hope this will help you...