Question

16. In the figure given below, angle BAC = 90° and AD is perpendicular to hypotenuse BC. Prove that: AD2 = BD X DC.
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Answer 1

Step-by-step explanation:

In ABC

seg AD perpendicular Hypotenuse BC

In ABD and ADC

all angles are congruent ......(given)

hence ABD ~ADC

Therefore

$\frac{ad}{ dc} = \frac{bd}{ad}$ .....(c.s.s.t )

Therefore AD X AD =BD X DC

Therefore

${ad}^{2} = bd \: \times dc$

hence it is proved

Answer 2

PROVED

AD² = BD X DC.

Step-by-step explanation:

given : AD is perpendicular to hypotenuse BC

$\angle BAC = 90^\circ$

to prove : AD² = BD X DC.

proof :

AD is perpendicular to hypotenuse BC

$\angle BAC = 90^\circ$

so , these triangles ABD and CAD is similar

$\bigtriangleup ABD \sim \bigtriangleup CAD$

by corresponding parts of similar triangles

if two triangles are similar then the ratio of the area of both the triangles is proportional to the square of the ratio of their corresponding sides

therefore ,

$\frac{BD}{AD}= \frac{AD}{CD}\\\\BD \times CD = AD \times AD \\\\BD \times CD =AD^2$

HENCE PROVED

#Learn more :

In triangle ABC AD perpendicular BC ,if (AD) square= BD × DC then prove that angle BAC =90

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