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# □ show : d² = ab - st.?

△ABC.

bisector of ∠C intersects AB at P.

BC = a , AC = b , AP = s , BP = t , CP = d.

show : d² = ab - st.

### 1 Answer

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- 3 years agoFavorite Answer
•b/a=s/t

bt = as.....(1)

• (b²+d²-s²)/2bd = (d²+a²-t²)/2ad

...ab²+ad²-as² = bd² + ba² - bt²

...ad² - bd² = (ba²-ab²) + as.s - bt.t

...d²(a-b) = ab(a-b) +bt.s - as.t

...d²(a-b)= ab(a-b) - st(a-b)

...d² = ab - st

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• cot(arccot 3 + arccot 7 + arccot 13 + arccot 21)

• tan-¹[((1/3) + (1/7))/(1-(1/3)(1/7))] +

tan-¹[(1/13)+(1/21)/(1-(1/21)(1/13)]

= tan-¹(1/2) + tan-¹(1/8)

= tan-¹[(1/2)+(1/8)/(1-(1/2)(1/8))]

=tan-¹(10/15)

=cot-1(15/10)

cotcot-¹(15/10) = 15/10