In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC² + BD² = AD² + BC² [Hint: Produce AB and DC to meet at E.]

Solution:

Given, ABCD is a quadrilateral

Alos, ∠A + ∠D = 90°

We have to prove that AC² + BD² = AD² + BC².

We know that the sum of all the three interior angles of a triangle is always equal to 180°

In △AED,

∠A + ∠D + ∠E = 180°

∠E + 90° = 180°

∠E = 180° - 90°

∠E = 90°

In △ADE,

By pythagoras theorem,

AD² = AE² + DE² --------------------- (1)

In △BEC,

By pythagoras theorem,

BC² = BE² + CE² --------------------- (2)

Adding (1) and (2),

AD² + BC² = AE² + DE² + BE² + CE²

AD² + BC² = AE² + CE² + BE² + DE² --------------- (3)

In △AEC,

By pythagoras theorem,

AC² = AE² + CE² ------------ (4)

In △BED,

By pythagoras theorem,

BD² = BE² + DE² ------------- (5)

Substituting (4) and (5) in (3),

AD² + BC² = AC² + BD²

Therefore, it is proved that AC² + BD² = AD² + BC²

✦ Try This: In △ABC, ∠B = 90° and D is the midpoint of BC. Prove that AC² = AD² + 3CD²

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 6

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NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 12

In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC² + BD² = AD² + BC² [Hint: Produce AB and DC to meet at E.]

Summary:

In a quadrilateral ABCD, ∠A + ∠D = 90°. It is proved that AC² + BD² = AD² + BC² by Pythagoras theorem which states that in a right triangle the square of the hypotenuse is equal to the sum of squares of the other two sides

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