Answer:
It is given that AD is an altitude and AB = AC. The diagram is as follows:
(i) In ΔABD and ΔACD,
∠ADB = ∠ADC = 90°
AB = AC (It is given in the question)
AD = AD (Common arm)
∴ ΔABD ≅ ΔACD by RHS congruence condition.
Now, by the rule of CPCT,
BD = CD.
So, AD bisects BC
(ii) Again, by the rule of CPCT, ∠BAD = ∠CAD
Hence, AD bisects ∠A.
Step-by-step explanation:
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that(i) AD bisects BC
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that(i) AD bisects BC (ii) AD bisects ∠A...
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Verified answer
Answer:
It is given that AD is an altitude and AB = AC. The diagram is as follows:
(i) In ΔABD and ΔACD,
∠ADB = ∠ADC = 90°
AB = AC (It is given in the question)
AD = AD (Common arm)
∴ ΔABD ≅ ΔACD by RHS congruence condition.
Now, by the rule of CPCT,
BD = CD.
So, AD bisects BC
(ii) Again, by the rule of CPCT, ∠BAD = ∠CAD
Hence, AD bisects ∠A.
Step-by-step explanation:
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that(i) AD bisects BC
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that(i) AD bisects BC (ii) AD bisects ∠A...