Question

In ABC ACB=90°, CD is perpendicular to side AB and Seg CE is angle bisector of ACB. Prove AD/BD = AE^2/ BE^2​

Answer 1

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

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

$\therefore\frac{AE^{2}}{BE^{2}}=\frac{AD}{BD}\text{ (Proved)}$