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In triangle ABC, D and E are points on AB and AC respectively such that DE||BC and DE divides triangle ABC into $2$ parts of equal areas. Find the ratio of AD and BD.

I am unable to start this question. Hope somebody could provide some hint.

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  • $\begingroup$ Hint: show the triangles ABC and ADE are similar. $\endgroup$ – Gerry Myerson Sep 9 '13 at 10:12
  • $\begingroup$ but they are indeed similar. one parallel side, one same angle, two proportionate sides. But how would it lead me to solve the question? Hope you could guide further. $\endgroup$ – aarbee Sep 9 '13 at 10:14
  • $\begingroup$ Second hint: How do the areas of similar triangle relate, if you know the scaling factor? $\endgroup$ – Hagen von Eitzen Sep 9 '13 at 10:19
  • $\begingroup$ So, what does that say about a side of ADE, in relation to a corresponding side of ABC? $\endgroup$ – Gerry Myerson Sep 9 '13 at 10:26
  • $\begingroup$ Let AG be the altitude of ABC, cutting DE at F. So, $\frac12*BC*AG=2*\frac12*DE*AF$. I am afraid it is not giving me anything. Certainly, $BC\ne2DE$, because had that been so, area of ADE would have been half that of trapezium DECB. $\endgroup$ – aarbee Sep 9 '13 at 10:35
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First note that that the the triangles $ADE$ and $ABC$ are simular, because they have 3 parallel sides.

For two simular triangle the following statements hold:

$$\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1} = \frac{h}{h_1} = k$$

$$\frac{P}{P_1} = k^2$$

Because the area of the smaller triangle is half of the bigger one we have:

$$\frac{2P}{P} = k^2$$ $$k^2 = 2$$ $$k = \sqrt{2}$$

Now from the first statement we have:

$$\overline{AB} = \overline{AD} \cdot k$$ $$\overline{AD} + \overline{BD} = \overline{AD} \cdot \sqrt{2}$$ $$\overline{BD} = (\sqrt{2} - 1)\overline{AD}$$ $$\frac{\overline{BD}}{\overline{AD}} = \sqrt{2} - 1$$ $$\frac{\overline{AD}}{\overline{BD}} = \frac{1}{\sqrt{2} - 1}$$

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  • $\begingroup$ what are the smaller letters a b and c? what is P represent? is the h height? k looks like a constant! great please explain the relationships $\endgroup$ – IshmaelR Mar 6 '14 at 21:58

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