if ABC be a triangle right angled at C if D is the midpoint of BC prove that AB2
= 4 AD2
- 3AC2
.
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Answer:In ∆ACB and ∆ACD
Pythagoras thm
AD²=DC²+AC²
Hence, AC²=AD²-DC²________(1)
NOW,
AB²=AC²+BC²
AB²=(AD²-DC²)+BC²_____FROM(1)
AB²=AD²-(½BC)²+BC²______D is midpoint
AB²=AD²-¼BC²+BC²
Multipy by 4
4AB²=4AD²-DC²+4BC²
3AB²+AB²=4AD²+3BC²
AB²=4AD²+3BC²-3AB²
AB²=4AD²-3(BC²+AB²)
AB²=4AD²-3AC²_____Pythagoras thm
Hence Proved.
Another soluation : Given: ABC be a triangle, right - angled at C and D is the mid - point of BC.
To Prove: AB2 = 4AD2 - 3AC.
Proof:
From right triangle ACB, we have,
AB2 = AC2 + BC2
= AC2 + (2CD)2 = AC2 + 4CD2 [since BC = 2CD]
= AC2 + 4(AD2 - AC2) [From right △ACD]
= 4AD2 - 3AC2.
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