Question

31. In AABC, AD 1 BC and BD-3 CD. Prove that 2 AB2-2 AC2+ BC2.
32. A cone of maximum​

Answer 1

Answer:

go through the attached sheet

Step-by-step explanation:

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Answer 2

We are given here

AD ⊥ BC

& DB=3CD −−−(1)

so, BC=DB+CD

BC=3CD+CD

so, BC=4CD −−−(2)

Hence it given that, AB is perpendicular to BC by the pythagoras theorem.

In both triangle △ABD & △ACD

AB² =AD² +BD² & AC² =AD² +CD²

∴ AD² =AB² −DB² & AD² =AC² −CD²

Compare both AD² with each other we get,,

AB² −DB² =AC² − CD²

AB² =AC² +DB² −CD²

AB² =AC² +(3 CD)² −CD² (from (1))

AB² =AC² +9 CD² −CD²

AB² =AC² +8 CD²

AB² = AC² +8(BC/4)² (from (2))

2AB² = 2AC² +BC²