Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides
Answers
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Let us consider that :
∠ABC is a right angle.
AC is the hypotenuse.
AB is the perpendicular.
BC is the base.
Given: In ΔABC, ∠ABC=90°
Construction: BD is a perpendicular on side AC
To Prove: (AC)²=(AB)²+(BC)²
Proof:
In △ABC,
∠ABC=90° (Given)
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.
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