Question

In isosceles triangle abc,ab=ac and a is joined to the mid point d of the side bc prove that ad perpendicular to bc and ad bisects the angle a​

Answer 1

Answer:

Since, the triangle is an isosceles with AB=AC

∠B = ∠C       ( ∴ The angles opposite to equal sides are always equal )

Now, BD = DC  ( ∴ D is the mid point  )

Now, from triangle ABD and triangle ACD,

AB=AC         (given)

∠B = ∠C      ( Proved above)

BD = CD       ( proved above )

Δ ABD ≅ Δ ACD        ( by SAS congruence rule)

Now,

∠BAD = ∠CAD       ( by CPCT )

AD bisects ∠A.

Again,

∠ADC = ∠ADB    ( ∴ Angles opposite to equal sides AB and AD )

Now,

∠ADB+∠ADC = 180°          ( Linear pair of angles )

Let each equal angles ∠ADB and ∠ADC be x then,

x+x = 180°  

2x=180°

x=180° /2

x=90°

∠ADB=∠ADC = 90°

AD⊥BC

Hence proved

Answer 2

CRM

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