Question

18. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects A.​

Answer 1

Answer:

$\huge\underbrace\mathfrak\pink{♡Answer♡}$

• Show that (i) AD bisects BC , (ii) AD bisects ∠A. Given: ∆ ABC is an isosceles triangle, So, AB = AC Also, AD is the altitude So, ∠ADC = ∠ADB = 90∘ To prove: (i) BD = CD & (ii) ∠BAD = ∠CAD Proof.

Answer 2

Given -

AD is an altitude

AB = AC

To find -

AD bisects BC

AD bisects ∠ A

Solution -

In Triangle ABD & Triangle ACD

AB = AC [given]

∠ B = ∠ C [angles opposite to the equal sides of a Triangle are equal]

AD = AD [common]

∴ Triangle ABD is congurent to Triangle ACD [ASA rule]

So,

AD Bisects BC [CPCT]

AD Bisects ∠ A [CPCT]