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Question
In ∆ABC, ∠ABC = 135°. Prove that AC2 = AB2 + BC2 + 4 ar (∆ABC)
Solution
We have the following figure.
Let Draw an altitude AD on Extended BC
In ∆ADC, ∠D = 90°
Therefore, by Pythagoras theorem, we have
`AC^2=AD^2+DC^2`
`AC^2=AD^2+(DB+BC)^2`
`AC^2=AD^2+DB^2+BC^2+2.BC.BD` ...(i)
In ΔADB, ∠D = 90°
⇒ `AD^2 + BD^2 = AB^2`
By substituting values of equation (i)
⇒`AD^2+DB^2+BC^2+2.BC.BD = AC^2`
⇒`AB^2 + BC^2 + 2BD.BC = AC^2`
As in ΔADB, ∠DAB = 45°
⇒ AD = DB (Opposite sides are equal)
⇒`AB^2+BC^2+2.BC.BD = AC^2`
⇒`AB^2+BC^2+2.BC.AD = AC^2`
⇒`AB^2+BC^2 + 4 xx 2/4 xx AD xx BC= AC^2`
∴ `1/2 xx AD xx BC = ar(ΔABC)`
⇒ `AC^2 = AB^2 + BC^2 + 4 ar (∆ABC)`
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