Question

ɪɴ ᴀ ʀɪɢʜᴛ ᴛʀɪᴀɴɢʟᴇ , ᴘʀᴏᴠᴇ ᴛʜᴀᴛ ᴛʜᴇ sǫᴜᴀʀᴇ ᴏғ ᴛʜᴇ ʜʏᴘᴏᴛᴇɴᴜsᴇ ɪs ᴇǫᴜᴀʟ ᴛᴏ ᴛʜᴇ sᴏᴍᴇ ᴏғ ᴛʜᴇ sǫᴜᴀʀᴇs ᴏɴ ᴛʜᴇ ᴏᴛʜᴇʀ ᴛᴡᴏ sɪᴅᴇ..

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Answer 1

$\huge\bold\green{Answer:-}$

Given:

A right angled ∆ABC, right angled at B.

To Prove- AC²=AB²+BC²

Construction: draw perpendicular BD onto the side AC .

Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

△ADB∼△ABC. (by AA similarity)

Therefore, AD/ AB=AB/AC

(In similar Triangles corresponding sides are proportional)

AB²=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CD/BC=BC/AC

(in similar Triangles corresponding sides are proportional)

Or, BC²=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( From the figure AD + CD = AC)

AB²+BC²=AC . AC

Therefore, AC²=AB²+BC²

This theroem is known as Pythagoras theroem...

[Figure provided in attachment]

Answer 2

Given :

A right angled triangle ABC with ∠B = 90°

To prove :

AC² = AB² + BC² (Pythagoras theorem)

Theorem used :

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and also to each other.

Construction :

Draw BD ⊥ AC

Proof :

Now, ΔADB ∼ ΔABC (by theorem)

So, $\dfrac{AD}{AB}$ = $\dfrac{AB}{AC}$ (as the sides are proportional)

=> AD. AC = AB² ------------(I)

And, ΔBDC ∼ ΔABC (by theorem)

So, $\dfrac{CD}{BC}$ = $\dfrac{BC}{AC}$ (as the sides are proportional)

=> CD. AC = BC² ------------(2)

Adding (1) and (2)

=> AD. AC + CD. AC = AB² + BC²

=> AC (AD + CD) = AB² + BC²

(∵ AD + CD = AC)

=> AC. AC = AB² + BC²

=> AC² = AB² + BC²

Hence proved !!