In ∆ABC, ∠C is a right angle. Points P & Q lies on the sides CA & CB respectively Prove that AQ² + BP² = AB² + PQ²
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PYTHAGORAS THEOREM: In a right angle triangle the square of the hypotenuse is equal to the sum of the square of the other two sides.
Given:
In right angled ∆ACB ,∠C = 90°, two Points P & Q lies on the sides CA & CB. Join PQ, AQ & PB.
To Prove:AQ² + BP² = AB² + PQ².
PROOF:
In right angled ∆ACQ,
AQ² = AC² + CQ²……………….(1)
[by Pythagoras theorem]
In right angled ∆PCB,
PB² = PC² + BC²……………….(2)
[by Pythagoras theorem]
In right angled ∆ACB,
AB² = AC² + CB²……………….(3)
[by Pythagoras theorem]
In right angled ∆PCQ,
PQ² = PC² + CQ²……………….(4)
[by Pythagoras theorem]
On Adding eq 1 & 2
AQ² + PB² = (AC² + CQ²) + (PC² + BC²)
AQ² + PB² = (AC² + BC²) + (PC² + CQ²)
AQ² + PB² = AB² + PQ²
[ From eq 3 & 4]
Hence, proved.
HOPE THIS WILL HELP YOU...
Given:
In right angled ∆ACB ,∠C = 90°, two Points P & Q lies on the sides CA & CB. Join PQ, AQ & PB.
To Prove:AQ² + BP² = AB² + PQ².
PROOF:
In right angled ∆ACQ,
AQ² = AC² + CQ²……………….(1)
[by Pythagoras theorem]
In right angled ∆PCB,
PB² = PC² + BC²……………….(2)
[by Pythagoras theorem]
In right angled ∆ACB,
AB² = AC² + CB²……………….(3)
[by Pythagoras theorem]
In right angled ∆PCQ,
PQ² = PC² + CQ²……………….(4)
[by Pythagoras theorem]
On Adding eq 1 & 2
AQ² + PB² = (AC² + CQ²) + (PC² + BC²)
AQ² + PB² = (AC² + BC²) + (PC² + CQ²)
AQ² + PB² = AB² + PQ²
[ From eq 3 & 4]
Hence, proved.
HOPE THIS WILL HELP YOU...
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