Question

QUESTION 10. !! With Steps
50 pts.

Answer 1

HELLO DEAR,









Given: ABC Is A Right ∆ Right Angled At A

And AD ⊥ BC

 

NOW In ∆ABC

∠A + ∠B + ∠C = 180° (Angle sum properly)

=>∠B + ∠C = 180° – ∠A

= 180° – 90° = 90°  ........... (1)

 

In ∆ABD

∠B + ∠BAD  + ∠ADB = 180°

=>∠B + ∠BAD = 180° – ∠ADB

= 180° – 90° = 90°  ........... (2)

 

In ∆ADC

=>∠DAC + ∠ADC + ∠C = 180°

=>∠C + ∠DAC = 180° – ∠ADC

= 180° – 90° = 90°  ........... (3)

 

NOW In ∆BAD and ∆ACD

∠BAD = ∠C  (from (1) and (2))

∠B = ∠DAC  (from (1) and (3))

∆BAD ~ ∆ACD (by AA similarly centers)

$\frac{bd}{ad} = \frac{ad}{cd}$


Thus BD : AD : : AD : CD


I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Answer:

A right triangle ABC right-angled at A,

TO PROVE AD2 = BD CD

Since ABD and ACD are right triangles.

AB2 = AD2 + BD2 (i)

and, AC2 = AD2 + CD2 (ii)

Adding equations (i) and (ii), we get

AB2 + AC2 = 2AD2 + BD2 + CD2

BC2 = 2 AD2 + BD2 + CD2 [ ABC is right-angled at A AB2 + AC2 = BC2]

(BD + CD)2 = 2AD2 + BD2 + CD2

BD2 + CD2 + 2 BD CD = 2 AD2 + BD2 + CD2

2BD CD = 2 AD2

AD2 = BD CD

Hence, AD2 = BD CD