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:
$AC^2= AD^2 + DC^2$
$\Rightarrow AC^2 - DC^2 = AD^2$
$\Rightarrow AD^2 = AC^2- DC^2 .....(1)$
Applying Oythagoras theorem in right-triangle ADB, we get:
$AB^2 = AD^2 + DB^2$
$\Rightarrow AB^2 - DB^2= AD^2$
$\Rightarrow AD^2 = AB^2 - DB^2 .....(2)$
From equation (1) and (2), we have:
$AC^2 - DC^2 = AB^2 - DB^2$
$\Rightarrow AC^2 = AB^2 + DC^2 - DB^2$
$\Rightarrow AC^2 = AB^2 + (DB + BC)^2- DB^2$ $(\therefore\text{DB}+\text{BC}=\text{DC})$
$\Rightarrow AC^2 = AB^2 + DB^2 + BC^2 + 2DB.BC - DB^2$
$\Rightarrow AC^2 = AB^2 + BC^2 + 2BC.BD$
This completes the proof.
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