Question
In $\triangle\text{ABC},$ given that $AB = AC$ and $\text{BD}\perp\text{AC}.$ Prove that $BC^2 = 2AC \times CD.$
✓
Answer
Given: In $\triangle\text{ABC},$
$\text{AB} = \text{AC, } \text{BD}\perp\text{AC}$
To prove: $BC^2 = 2AC \times CD$
$\therefore$
$AB^2 = BD^2 + AD^2 $(Pythagoras Theorem)
$\Rightarrow BD^2 = AB^2 - AD^2 .....(i)$ Similarly in right
$\triangle\text{BDC},$$ BC^2 = BD^2 + DC^2 = AB^2 - AD^2 + DC^2$^ [From (i)]
$= AB^2 - (AC - CD)^2 + CD^2 $
$= AB^2 - (AC^2 + CD^2 - 2AC \times CD) + CD^2 $
$= AC^2 - AC^2 - CD^2 + 2AC \times CD + CD^2$^
($\because$
$AB = AC) = 2AC \times CD$ (given)
Hence $BC^2 = 2AC \times CD$
$\text{AB} = \text{AC, } \text{BD}\perp\text{AC}$
To prove: $BC^2 = 2AC \times CD$
$\therefore$
$AB^2 = BD^2 + AD^2 $(Pythagoras Theorem)
$\Rightarrow BD^2 = AB^2 - AD^2 .....(i)$ Similarly in right
$\triangle\text{BDC},$$ BC^2 = BD^2 + DC^2 = AB^2 - AD^2 + DC^2$^ [From (i)]
$= AB^2 - (AC - CD)^2 + CD^2 $
$= AB^2 - (AC^2 + CD^2 - 2AC \times CD) + CD^2 $
$= AC^2 - AC^2 - CD^2 + 2AC \times CD + CD^2$^
($\because$
$AB = AC) = 2AC \times CD$ (given)
Hence $BC^2 = 2AC \times CD$
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