Question
In $\triangle\text{ABC},$ if $\text{BD}\perp\text{AC}$ and $BC^2 = 2AC \times CD$, then prove that $AB = AC.$
✓
Answer
Since $\triangle\text{ADB}$ is right triangle right angled at $D$
$AB^2 = AD^2 + BD^2$
In right $\triangle\text{BDC,}$ we have
$CD^2 + BD^2 = BC^2$
Since $2AC \times DC = BC^2$
$\Rightarrow DC^2 + BD^2 = 2AC \times DC$
$2AC \times DC = AC^2 - AC^2 + DC^2 + BD^2$
$AC^2 = AC^2 + DC^2 - 2AC \times DC + BD^2$
$AC^2 = (AC - DC)^2 + BD^2$
$AC^2 = AD^2 + BD^2$
Now substitute $AD^2 + BD^2 = AB^2$
$AC^2 = AB^2$
$AC = AB$
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