Question
In the given figure, $\triangle\text{ABC}$ is an obtuse triangle, obtuse-angled at B. If $\text{AD}\perp\text{CB}$ (produced) prove that $AC^2= AB^2+ BC^2+2BC.BD.$
✓
Answer
Applying Pythagoras theorem in right-angled triangle $ADC$, we get:
$A C^2=A D^2+D C^2 $
$ \Rightarrow A C^2-D C^2=A D^2 $
$ \Rightarrow A D^2=A C^2-D C^2...(1)$
Applying Oythagoras theorem in right-triangle $ADB$, we get:
$ A B^2=A D^2+D B^2 $
$ \Rightarrow A B^2-D B^2=A D^2 $
$ \Rightarrow A D^2=A B^2-D B^2...(2)$
From equation $(1)$ and $(2)$,, we have:
$A C^2-D C^2=A B^2-D B^2 $
$ \Rightarrow A C^2=A B^2+D C^2-D B^2 $
$ \Rightarrow A C^2=A B^2+(D B+B C)^2-D B^2(\therefore D B+B C=D C) $
$ \Rightarrow A C^2=A B^2+D B^2+B C^2+2 D B \cdot B C-D B^2 $
$ \Rightarrow A C^2=A B^2+B C^2+2 B C \cdot B D$
This completes the proof.
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