In a right triangle ABC, right-angled at B, D is a point on hypotenuse such that BD . If DP and DQ then prove that. 1. DQ^2=DP.DQ 2. DP^2=DQ.AP

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. 1. In , DQ DQ = QB QC ^2=QB.QC Since all the angles of PBQD are 90^ , PBQD is a rectangle. QB = DP and PB = DQ ....(i) DQ^2 = DP . QC 2. Similarly, since PD is a perpendicular on AB, DP PB = AP DP DP DQ = AP DP (using (i)) ^2=DQ.AP

In a right triangle ABC, right-angled at B, D is a point on hypot…

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Question

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.

$\text{DQ}^2=\text{DP}.\text{DQ}$
$\text{DP}^2=\text{DQ}.\text{AP}$

✓

Answer

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.

In $\triangle\text{DBC},$

$\triangle\text{DQB}\sim\triangle\text{DQC}$
$\Rightarrow\frac{\text{DQ}}{\text{DQ}}=\frac{\text{QB}}{\text{QC}}$
$\Rightarrow\text{DQ}^2=\text{QB}.\text{QC}$
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 $AB$,

$\triangle\text{APD}\sim\triangle\text{DPB}$
$\Rightarrow\frac{\text{DP}}{\text{PB}}=\frac{\text{AP}}{\text{DP}}$
$\Rightarrow\frac{\text{DP}}{\text{DQ}}=\frac{\text{AP}}{\text{DP}}\dots(\text{using }(\text{i}))$
$\Rightarrow\text{DP}^2=\text{DQ}.\text{AP}$