Question

Fig. 7.39
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC
(ii) AD bisects Z A.
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Answer 1

Hello, Buddy!!

ɢɪᴠᴇɴ:-

• ABC is an Isosceles Triangle.
• AD is the altitude of ∆ABC.
• AB = AC.

ᴛᴏ ᴘʀᴏᴠᴇ:-

• AD bisects BC.
• AD bisects ∠BAC.

ʀᴇQᴜɪʀᴇᴅ ꜱᴏʟᴜᴛɪᴏɴ:-

As,

AD is Perpendicular to BC [AD is the altitude of ∆ABC]

☞ ∠ADB = ∠ADC → 90°

WKT

Angles Opposite To Equal Sides are Equal.

As, AB = AC

☞ ∠ABC = ∠ACB

In ∆ADB & ∆ADC

AB = AC [Given]

∠ABC = ∠ACB [Proved Above]

∠ADB = ∠ADC [AD⊥BC]

∆ABD = ∆ACD [By ASA Axiom]

By CPCT

☞ BD = BC

Hence, We can say that AD bisects BC.

☞ ∠DAC = ∠DAB

Hence, We can say that AD bisects ∠BAC.

• Hence, Proved!!

@MrMonarque♡

Hope It Helps You ✌️