Question

The perpendicular AD on the base BC of a A ABC intersects BC at D
so that DB = 4 CD. Prove that 5AB2 = 5AC² + 3BC²​

Answer 1

Answer:

02-Sep-2017 · 2 answers

In triangle ABC, AD perpendicular to BC and point D lies on ... Prove that 5AB²=5AC²+BC² - 1441211. ... 2DB =3 CD, we know BC = BD + CD BC = 2/3 DB+ DB And DB ...

Answer 2

Answer:

Since ⊿ADB is a right triangle, we have

AB² = AD² + DB².

And since 2DB =3 CD, we know BC = BD + CD BC = 2/3 DB+DB  And DB = (3/5) CB.

(1) AB² = AD² + 9/25 BC2.

Similarly ⊿ADC is a right triangle, so

AC² = AD² + DC²,

So Similarly, DC = BC - BD DC= BC (2/5), and

(2) AC² = AD² + (4/25) BC²,

Subtract (1) by (2)

AB² - AC² = (9 - 4)/25 BC² AB² - AC² = 1/5 BC² 5AB² - 5AC² = BC²

So

5 AB² = 5 AC² + BC².

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