Question

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Answer 1

Answer:

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Step-by-step explanation:

Proof of the Pythagoras Theorem using Similarity of Triangles:

Given: In ΔABC, m∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

In △ABC,

m∠ABC=90° (Given)

seg BD is perpendicular to hypotenuse AC (Construction)

Therefore, △ADB∼△ABC∼△BDC (Similarity of right-angled triangle)

△ABC∼△ADB

(AB/AC)=(AD/AB) (congruent sides of similar triangles)

AB2=AD×AC (1)

△BDC∼△ABC

CD/BC=BC/AC (congruent sides of similar triangles)

BC2=CD×AC (2)

Adding the equations (1) and (2),

AB2+BC2=AD×AC+CD×AC

AB2+BC2=AC(AD+CD)

Since, AD + CD = AC

Therefore, AC2=AB2+BC2

Hence Proved.

Answer 2

Given,,,,

➡ ∆ABC in which ....angle ABC = 90°

◾To prove:-

➡ AC² = AB²+BC²

◾Construction:-

➡ Draw BD perpendicular AC...

◾Proof:-

➡ In ∆ABD and ∆ABC ,,we have

Angle A = Angle A (common)

Angle ABD = Angle ABC (equal to 90°)

Therefore, ∆ADB ~ ∆ABC (By AA similarity)

AD / AB = AB / BC

➡ AB² = AD × AC --------> (eq.1)

◾In ∆BDC and ∆ABC ,,, we have

Angle C = Angle C (common)

Angle BDC and Angle ABD (equal to 90°)

Therefore,,, ∆BDC ~ ∆ABC (By AA similarity)

➡DC / BC = BC / AC

➡ BC² = DC × AC --------> (eq.2)

◾Adding eq. 1 and 2 ,,, we get

➡ AB²+BC² = AD × AC + DC × AC

➡ AB²+BC² = AC (AD+DC)

➡ AB² + BC² = AC × AC

◾AC² = AB²+ BC²

Hence,,, proved...