In the given figure, ABD is a triangle right angled at A and $AC \perp BD$. Show that $AD^2= BD.CD$
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AcademicMathematicsNCERTClass 10

Given:

$\triangle ABD$ is a right triangle right-angled at A and $AC \perp BD$. 

To do: 

We have to show that $AD^2=BD.CD$.

Solution:

In $\triangle DCA$ and $\triangle DAB$,

$\angle DCA=\angle DAB=90^o$

$\angle CDA=\angle BDA$    (Common angle) 

Therefore,

$\triangle DCA \sim\ \triangle DAB$   (By AA similarity)

This implies,

$\frac{DC}{DA}=\frac{DA}{DB}$   (Corresponding parts of similar triangles are proportional)

$DA.DA=DB.DC$   (By cross multiplication)

$AD^2=BD.CD$

Hence proved. 

raja
Updated on 10-Oct-2022 13:21:26

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