Draw that in a right triangle square of hypatonus is equal to sum of the equal of an the other side
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Given: A right triangle ABC right angled at B.
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC
Proof:
In Δ ADB and Δ ABC,
∠ ADB = ∠ ABC (each 90°)
∠ BAD = ∠ CAB (common)
Δ ADB ~ Δ ABC (By AA similarity criterion)
Now, AD/AB = AB/AC (corresponding sides are proportional)
AB2 = AD × AC … (i)
Similarly, Δ BDC ~ Δ ABC
BC2 = CD × AC … (ii)
Adding (i) and (ii)
AB2 + BC2 = (AD × AC) + (CD × AC)
AB2 + BC2 = AC × (AD + CD)
AB2 + BC2 = AC2
Hence Proved.
To Prove: AC2 = AB2 + BC2
Construction: Draw BD ⊥ AC
Proof:
In Δ ADB and Δ ABC,
∠ ADB = ∠ ABC (each 90°)
∠ BAD = ∠ CAB (common)
Δ ADB ~ Δ ABC (By AA similarity criterion)
Now, AD/AB = AB/AC (corresponding sides are proportional)
AB2 = AD × AC … (i)
Similarly, Δ BDC ~ Δ ABC
BC2 = CD × AC … (ii)
Adding (i) and (ii)
AB2 + BC2 = (AD × AC) + (CD × AC)
AB2 + BC2 = AC × (AD + CD)
AB2 + BC2 = AC2
Hence Proved.
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