In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that : 2AC² - AB² = BC² + CD² + DA²
Answers
Answered by
1
Answer:
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Step-by-step explanation:
Given ∠B=∠D=90
∘
So ΔABC and ΔADC are right triangles.
In ΔABC,
AC
2
=AD
2
+BC
2
...(i) [Pythagoras theorem]
ΔADC,
AC
2
=AD
2
+DC
2
...(ii) [Pythagoras theorem]
Adding (i) and (ii)
2AC
2
=AB
2
+BC
2
+AD
2
+DC
2
∴2AC
2
−BC
2
=AB
2
+AD
2
+DC
2
Hence proved.
Answered by
0
Answer:
Proved
2AC²-AB²=BC²+CD²+DA²
Step-by-step explanation:
Rewrite the given"
Given,
<B = _D = 90°
SoAABC and AADC are right triangles.
In AABC
Solving the equation
Delta*ADC
A * C ^ 2 = A * D ^ 2 + D * C ^ 2
2AC^ 2 = A * B ^ 2 + B * C ^ 2 + A * D ^ 2 + D * C ^ 2
2A * C ^ 2 - B * C ^ 2 =AB^ 2 +AD^ 2 + D * C ^ 2
Final answer
If the quadrilateral value is 2A * C ^ 2 - B * C ^ 2 = A * B ^ 2 + A * D ^ 2 + D * C ^ 2 hence it is proved.
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