Question

In the adjoining figure, ABC is a triangle in which AD is the bisector of ∠A. If AD ⊥ BC, show that ∆ABC is isosceles.​

Answer 1

Answer:

Step-by-step explanation:

Given : AD bisects ∠A and is perpendicular to BC.

To prove : ΔABC is isoceles

Proof :

In ΔABD & ΔADC

∠BAD=∠DAC            (∵  D bisects ∠A)

∠ADB=∠ADC

⇒∠ABD=∠ACD

(∵  if two angles in a triangle are equal to two angles in another triangle, then the third angle of both the triangles must also be equal)

⇒∠B=∠C

∴   AB=AC (if two angles are equal then the side opposite to the angles will also be equal)

Hence proved.

Answer 2

Answer:

In Triangle ADB and Triangle ADC,

Angle ADB = Angle ADC (90°)

BD = DC (D is mid-point and AD|BC)

AD = AD (Common)

So, Triangle ADB is congruent to Triangle ADC

AB = AC (CPCT)

So, Triangle ABC is isoceles.

Hope It Helps ✌✌