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In ΔАВC, ∠ABC = 90° and D is any point on BC. Prove that : AD^2 + BC^2 = AC^2 + BD^2. - Mathematics

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Question

In ΔАВC, ∠ABC = 90° and D is any point on BC. Prove that : AD2 + BC2 = AC2 + BD2.

Theorem
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Solution

Given:

In triangle ABC, ∠ABC = 90°.

D is any point on BC.

To Prove:

AD2 + BC2 = AC2 + BD2.

Proof [Step-wise]:

1. Since ∠ABC = 90° and D lies on BC, BA ⟂ BD. 

So, triangle ABD is right-angled at B.

By the Pythagorean theorem in ΔABD: 

AD2 = AB2 + BD2

2. Also, triangle ABC is right-angled at B.

By the Pythagorean theorem in ΔABC:

AC2 = AB2 + BC2

3. Subtract the second equation from the first:

AD2 – AC2

= (AB2 + BD2) – (AB2 + BC2

= BD2 – BC2

4. Rearranging gives

AD2 + BC2 = AC2 + BD2

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Chapter 10: Pythagoras Theorem - Exercise 10A [Page 211]

APPEARS IN

Nootan Mathematics [English] Class 9 ICSE
Chapter 10 Pythagoras Theorem
Exercise 10A | Q 17. | Page 211
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