Question

18. Show that ∆ABD = ∆ACD.
Hence prove that ∆ADB = ∆ADC= 90° ​

Answer 1

Answer:

they are equal

Step-by-step explanation:

in triangle ABD and ADC:-

AB=AC(given)

BD=DC(given)

AD=AD(common)

so, by sss rule

they both are equal

Also given, BD=DC

so, AD bisects BC

because of AD perpendicular to BC

therefore , Angle ADB and ADC=90°

Answer 2

Answer:

HEY .....(^_^)

√3 + √2 / √3 - √2

= (√3 + √2 / √3 - √2 ) × ( √3 + √2 / √3 + √2)

= (√3 + √2)^2 / (√3)^2 - (√2)^2

= [ (√3)^2 + (√2)^2 + 2×√3×√2 ] / 3-2

= 3 + 2 + 2√6 / 1

= 5 + 2√6

Step-by-step explanation: