In ΔABC, AD is the perpendicular from A to BC, with D lying on BC. Which relation is always true?

Class 10Triangles (Similarity)Geometry — Similarity

Step-by-Step Solution

Pythagoras in the two right triangles gives

A B^{2} = A D^{2} + B D^{2}

and

A C^{2} = A D^{2} + C D^{2}

. Subtracting removes A

D^{2} : A B^{2} - B D^{2} = A C^{2} - C D^{2}

, which rearranges to

A B^{2} + C D^{2} = A C^{2} + B D^{2}

. The first option pairs each side with its OWN foot-segment, which is what the subtraction shows to be false in general.

Answer: B)

A B^{2} + C D^{2} = A C^{2} + B D^{2}

**Theorem**: In a right triangle with legs a,

b

and hypotenuse

c : a^{2} + b^{2} = c^{2}

.

**Proof using similar triangles**:
In Δ

A BC

right-angled at

C

, drop perpendicular from

C

A B

meeting it at

D

.
- Δ

A C D

~ Δ

A BC (AA

: shared ∠A, both right-angled).
- Δ

BC D

~ Δ

A BC (AA

: shared ∠

B

, both right-angled).
From Δ

A C D

~ Δ

A BC : A C^{2} = A D \times A B

.
From Δ

BC D

~ Δ

A BC : B C^{2} = B D \times A B

.
Add:

A C^{2} + B C^{2} = A B (A D + B D) = A B \cdot A B = A

B^{2}

.
So

a^{2} + b^{2} = c^{2}

. ∎

**Converse**: if

a^{2} + b^{2} = c^{2}

in a triangle, then the triangle is right-angled at the vertex opposite

c

.

**Pythagorean triples**: integer triples (a,

b

, c) with

a^{2} + b^{2} = c^{2}

. Examples: (3, 4, 5), (5, 12, 13), (8, 15, 17), (7, 24, 25).

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