In a right triangle prove that the square of the hypotenuse
is equal to the sum of the squares of the other two sides.
Answers
Step-by-step explanation:
Given:- A right triangle ABC right angled at B.
To prove:- AC
2
=AB
2
+BC
2
Construction:- Draw BD⊥AC
Proof:-
In △ABC and △ABD
∠ABC=∠ADB(Each 90°)
∠A=∠A(Common)
∴△ABC∼△ABD(By AA)
AC
AB
=
AB
AD
(∵Sides of similar triangles are proportional)
⇒AB
2
=AD⋅AC.....(1)
Similarly, in △ABC and △BCD
∠ABC=∠BDC(Each 90°)
∠C=∠C(Common)
∴△ABC∼△BCD(By AA)
∴
BC
DC
=
AC
BC
⇒BC
2
=DC⋅AC.....(2)
Adding equation (1)&(2), we have
AB
2
+BC
2
=AD⋅AC+DC.AC
⇒AB
2
+BC
2
=AC(AD+DC)
⇒AB
2
+BC
2
=AC
2
Hence proved.
solution
Given:- A right triangle ABC right angled at B.
To prove:- AC²=AB²+BC²
Construction:- Draw BD⊥AC
Proof:-
In △ABC and △ABD
∠ABC=∠ADB(Each 90°)
∠A=∠A(Common)
∴△ABC∼△ABD(By AA)
ACAB =ABAD
(∵Sides of similar triangles are proportional)
⇒AB²=AD⋅AC.....(1)
Similarly, in △ABC and △BCD
∠ABC=∠BDC(Each 90°)
∠C=∠C(Common)
∴△ABC∼△BCD(By AA)
∴ BCDC=ACBC
⇒BC²=DC⋅AC.....(2)
Adding equation (1)&(2), we have
AB +BC²=AD⋅AC+DC.AC
⇒AB²+BC²=AC(AD+DC)
⇒AB²+BC²=AC²
Hence proved.