Question

prove converse of pythagoras theorem ​

Answer 1

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Answer 2

Answer:

Statement:In a right angle triangle the square of a hypotenuse is equal to the sum of the square of the other two sides

Given:ABC is a right angled triangle , right angled at b

prove that: AC^2 = AB^2 + BC^2

Construction:Draw BD perpendicular to AC

Step-by-step explanation:

Proof:In triangle ABD and triangle ABC

Angle A = Angle A \ Common angle

Angle B = Angle D = 90°

Therefore triangle ABD ~ triangle ABC \ A.A Criteria

AB/AD = AC/AB \ By CPCT

AB^2 = AD×AC ..................(1)

|||y In triangle BDC and triangle ABC

Angle C = Angle C \ Common angle

Angle B = Angle D = 90°

Therefore triangle BCD ~ triangle ABC \ A.A Criteria

BC/DC = AC/BC \ By CPCT

BC^2 = DC×AC ................(2)

From (1) & (2)

AB^2 + BC^2 = (AD×AC) + (DC×AC)

AB^2 + BC^2 = AC(AD + DC)

AB^2 + BC^2 = AC(AC)

AB^2 + BC^2 = AC^2

Hence proved