Question
$\triangle ABD$ is a right triangle right $-$ angled at $A$ and $AC \perp BD$. Show that $\frac{A B^2}{A C^2}=\frac{B D}{D C}$
✓
Answer
Given: $\triangle ABD$ is a right triangle right angled at $A$ and $AC \perp BD$.
To Prove: $\frac{A B^2}{A C^2}=\frac{B D}{D C}$
Proof: We know that if a perpendicular is drawn from the vertex of the right angle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
So, $\triangle B A D \sim \triangle B C A............(i)$
and $\triangle A C B \sim \triangle D C A.............(ii)$
If two triangles are similar, then the ratio of their corresponding sides are equal.
$\frac{B A}{B C}=\frac{B D}{B A}\ [$from $(i)]$
$BA^2=BC \times BD........(iii)$
Also, $\frac{A C}{D C}=\frac{B C}{A C} \ [$from $(ii)]$
$AC ^2= DC \times BC ........... (iv)$
Hence $\frac{A B^2}{A C^2}=\frac{C B \times B D}{D C \times B C}$
$\frac{A B^2}{A C^2}=\frac{B D}{D C}$
Hence proved.
To Prove: $\frac{A B^2}{A C^2}=\frac{B D}{D C}$
Proof: We know that if a perpendicular is drawn from the vertex of the right angle to the hypotenuse then the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.
So, $\triangle B A D \sim \triangle B C A............(i)$
and $\triangle A C B \sim \triangle D C A.............(ii)$
If two triangles are similar, then the ratio of their corresponding sides are equal.
$\frac{B A}{B C}=\frac{B D}{B A}\ [$from $(i)]$
$BA^2=BC \times BD........(iii)$
Also, $\frac{A C}{D C}=\frac{B C}{A C} \ [$from $(ii)]$
$AC ^2= DC \times BC ........... (iv)$
Hence $\frac{A B^2}{A C^2}=\frac{C B \times B D}{D C \times B C}$
$\frac{A B^2}{A C^2}=\frac{B D}{D C}$
Hence proved.
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