D and E are Points On the Sides CA and CB Respectively​ of a Triangle of a Traingal ABC Right Angled at C. Prove that AE^2+BD^2=AB^2+DE^2 | Bzziii.com

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D and E are points on the sides CA and CB respectively of a triangle of a traingal ABC right angled at C. Prove that `AE^2+BD^2`=`AB^2+DE^2`







Given, Triangle ABC, right angled at C

Two Points D and E are on the sides CA and CB

To Prove that, `"AE"_2` + `"BD"_2` = `"AB"_2` + `"DE"_2`

Let us join the points D with E and B and P with A

Using Pythagoras theorem

`"(Hypotenuse)"_2` = `"(Height)"_2` +`"(Base)"_2`


Here, ACE is a right angle triangle

`"AE"_2` = `"AC"_2` +`"CE"_2` .....(1)


In right angle triangle DCB

`"BD"_2` = `"DC"_2` +`"BC"_2` .....(2)


In right angle triangle ABC

`"AB"_2` = `"AC"_2` +`"BC"_2` .....(3)


In right angle triangle DCE

`"DE"_2` = `"DC"_2` +`"CE"_2` .....(4)


L.H.S

`"AE"_2` + `"BD"_2`

= `"(AC"_2` + `"CE"_2)` + `"(DC"_2` + `"BC"_2)`

= `"AC"_2` + `"CE"_2` + `"DC"_2` + `"BC"_2`
 R.H.S

`"AB"_2` + `"DE"_2`

= `"(AC"_2` + `"BC"_2)` + `"(DC"_2` + `"CE"_2)`

= `"AC"_2` + `"CE"_2` + `"DC"_2` + `"BC"_2`

Since, R.H.S = L.H.S

Hence, it is Proved that  `AE^2+BD^2`=`AB^2+DE^2`

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