Question : ABC is a right angle triangle and $\angle ABC = 90^{\circ}$. BD is perpendicular on the side AC. What is the value of $(BD)^2$?
Option 1: $AD\times DC$
Option 2: $BC\times AB$
Option 3: $BC\times CD$
Option 4: $AD\times AC$
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Correct Answer: $AD\times DC$
Solution :
Given: ABC is a right angle triangle and $\angle ABC = 90^{\circ}$.
In $\triangle ADB$ and $\triangle BDC$,
$\angle ADB=\angle BDC$ (Since $BD \perp AC$)
$\angle ABD=\angle BCD$
$BD = BD$ (common)
So, $\triangle ADB\sim\triangle BDC$
When two triangles are similar, their corresponding sides have the same proportion and their corresponding angles are congruent.
⇒ $\frac{AD}{DB}=\frac{BD}{DC}$
⇒ $(BD)^2=AD\times DC$
Hence, the correct answer is $AD\times DC$.
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