Question

4. AD is an altitude of an isosceles AABC in
which AB = AC.
→ Show that (i) AD bisects BC,
(ii) AD bisects ZA.​

Answer 1

Answer:

Given,

ABC is an isosceles triangle,

so AB = AC ....(1)

Also, AD is the altitude

so ∠ADC = ∠ADB = 90°

In ADB and ADC,

∠ADC = ∠ADB = 90°   (Both 90°)

AB = AC    (from(1))

AD = AD (common)

i.e., ADB = ADC  (RHS Congruency)

Hence by CPCT,

BD = DC and

∠BAC = ∠DAC

Hence proved :)

Answer 2

Step-by-step explanation:

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