Question

In ∆ABC , angle ABC = 90°. seg CD is perpendicular on side AB and seg CE is angle bisector of angle ACB.
Prove that : AD/BD= AE×AE/BE×BE.

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

As per the property of a angle bisector of a triangle, when an angle bisector is drawn in a triangle, the ratio of the opposite sides forming the bisected angle is equal to the ratio of the segments formed by bisector intersecting the opposite side.

In $\Delta ACB$

Seg CE bisects \angle ACB

$\therefore \frac{AE}{BE}=\frac{AC}{BC}$

Squaring the both side we get

$\frac{AE^{2}}{BE^{2}}=\frac{AC^{2}}{BC^{2}}\text{ .................equation-1}$

In $\Delta ABC$

$m\angle ACB=90^{\circ}$

Seg CD hypotenuse AB

As per the theorem on similarity of right angled triangles we know that when a perpendicular is drawn on the hypotenuse the triangle is divided into two parts, As the corresponding angles of this two triangles are equal so we can say that both triangles are similar.

therefore,

$\Delta ABC\sim\Delta ADC\sim\Delta BDC$

In $\Delta ABC$ and $\Delta ADC$

$\frac{AC}{AB}=\frac{AD}{AC}\\\\\Rightarrow AC^{2}=AB\times AD\text{ ..............equation-2}$

In $\Delta ABC$ and $\Delta BDC$

$\frac{BC}{AB}=\frac{BD}{BC}\\\\\Rightarrow BC^{2}=BD\times AB\text{ ..............equation-3}$

From equation - 1 we get

$\frac{AE^{2}}{BE^{2}}=\frac{AC^{2}}{BC^{2}}\\\\\Rightarrow\frac{AE^{2}}{BE^{2}}=\frac{AB\times AD}{BD\times AB}\\\\\Rightarrow\frac{AE^{2}}{BE^{2}}=\frac{AD}{BD}$

$\therefore\frac{AD}{BD}=\frac{AE\times AE}{BE\times BE}=\text{ (Proved)}$