# In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC^{2} + BD^{2} = AD^{2} + BC^{2} [Hint: Produce AB and DC to meet at E.]

**Solution:**

Given, ABCD is a quadrilateral

Alos, ∠A + ∠D = 90°

We have to prove that AC^{2} + BD^{2} = AD^{2} + BC^{2}.

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,

AD^{2} = AE^{2} + DE^{2} --------------------- (1)

In △BEC,

By pythagoras theorem,

BC^{2} = BE^{2 }+ CE^{2} --------------------- (2)

Adding (1) and (2),

AD^{2} + BC^{2 }= AE^{2} + DE^{2} + BE^{2} + CE^{2}

AD^{2} + BC^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2} --------------- (3)

In △AEC,

By pythagoras theorem,

AC^{2} = AE^{2} + CE^{2} ------------ (4)

In △BED,

By pythagoras theorem,

BD^{2} = BE^{2} + DE^{2} ------------- (5)

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

AD^{2} + BC^{2} = AC^{2} + BD^{2}

Therefore, it is proved that AC^{2} + BD^{2} = AD^{2} + BC^{2}

**✦ Try This: **In △ABC, ∠B = 90° and D is the midpoint of BC. Prove that AC^{2} = AD^{2} + 3CD^{2}

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

**NCERT Exemplar Class 10 Maths Exercise 6.4 Problem 12**

## In a quadrilateral ABCD, ∠A + ∠D = 90°. Prove that AC^{2} + BD^{2} = AD^{2} + BC^{2} [Hint: Produce AB and DC to meet at E.]

**Summary:**

In a quadrilateral ABCD, ∠A + ∠D = 90°. It is proved that AC^{2} + BD^{2} = AD^{2} + BC^{2} 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

**☛ Related Questions:**

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