In a right triangle $ABC,$ right-angled at $B, D$ is a point on hypotenuse such that $\text{BD}\perp\text{AC}.$ If $\text{DP}\perp\text{AB}$ and $\text{DQ}\perp\text{BC}$ then prove that.
- $\text{DQ}^2=\text{DP}.\text{DQ}$
- $\text{DP}^2=\text{DQ}.\text{AP}$
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We know that
when a perpendicular is drawn from the vertex of a triangle on to the hypotenuse, then the triangles on both sides of the perpendicular are similar to eachother and the to the whole triangle.
$\Rightarrow\frac{\text{DQ}}{\text{DQ}}=\frac{\text{QB}}{\text{QC}}$
$\Rightarrow\text{DQ}^2=\text{QB}.\text{QC}$
Since all the angles of PBQD are $90^\circ ,$
$PBQD$ is a rectangle.
$\Rightarrow QB = DP$ and $PB = DQ ....(i)$
$\Rightarrow DQ^2 = DP . QC$
$\Rightarrow\frac{\text{DP}}{\text{PB}}=\frac{\text{AP}}{\text{DP}}$
$\Rightarrow\frac{\text{DP}}{\text{DQ}}=\frac{\text{AP}}{\text{DP}}\dots(\text{using }(\text{i}))$
$\Rightarrow\text{DP}^2=\text{DQ}.\text{AP}$
when a perpendicular is drawn from the vertex of a triangle on to the hypotenuse, then the triangles on both sides of the perpendicular are similar to eachother and the to the whole triangle.
- In $\triangle\text{DBC},$
$\Rightarrow\frac{\text{DQ}}{\text{DQ}}=\frac{\text{QB}}{\text{QC}}$
$\Rightarrow\text{DQ}^2=\text{QB}.\text{QC}$
Since all the angles of PBQD are $90^\circ ,$
$PBQD$ is a rectangle.
$\Rightarrow QB = DP$ and $PB = DQ ....(i)$
$\Rightarrow DQ^2 = DP . QC$
- Similarly, since $PD$ is a perpendicular on $AB$,
$\Rightarrow\frac{\text{DP}}{\text{PB}}=\frac{\text{AP}}{\text{DP}}$
$\Rightarrow\frac{\text{DP}}{\text{DQ}}=\frac{\text{AP}}{\text{DP}}\dots(\text{using }(\text{i}))$
$\Rightarrow\text{DP}^2=\text{DQ}.\text{AP}$
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