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12. 如图，在△ABC中，DE // AC，EF // AB.
（1）求证：△BDE ∽ △EFC；
（2）若BD = $\frac{2}{3}$AD，且△BDE的面积是5，求△EFC的面积.
（1）求证：△BDE ∽ △EFC；
（2）若BD = $\frac{2}{3}$AD，且△BDE的面积是5，求△EFC的面积.
答案
12. (1) $\because DE // AC, \therefore ∠ BED = ∠ C. \because EF // AB, \therefore ∠ B = ∠ FEC. \therefore △ BDE ∽ △ EFC$
(2) $\because BD = \frac{2}{3}AD, \therefore \frac{BD}{AD} = \frac{2}{3}. \because DE // AC, \therefore \frac{BE}{CE} = \frac{BD}{AD} = \frac{2}{3}$. 由(1)知, $△ BDE ∽ △ EFC$, 且 $S_{△ BDE}=5. \therefore \frac{S_{△ BDE}}{S_{△ EFC}} = (\frac{BE}{CE})^2 = (\frac{2}{3})^2 = \frac{4}{9}. \therefore S_{△ EFC} = \frac{9}{4} S_{△ BDE} = \frac{9}{4} × 5 = \frac{45}{4}$
(2) $\because BD = \frac{2}{3}AD, \therefore \frac{BD}{AD} = \frac{2}{3}. \because DE // AC, \therefore \frac{BE}{CE} = \frac{BD}{AD} = \frac{2}{3}$. 由(1)知, $△ BDE ∽ △ EFC$, 且 $S_{△ BDE}=5. \therefore \frac{S_{△ BDE}}{S_{△ EFC}} = (\frac{BE}{CE})^2 = (\frac{2}{3})^2 = \frac{4}{9}. \therefore S_{△ EFC} = \frac{9}{4} S_{△ BDE} = \frac{9}{4} × 5 = \frac{45}{4}$
13. 如图，D、E分别是AC、AB上的点，$∠ ADE=∠ B$，$AG ⊥ BC$于点G，$AF ⊥ DE$于点F，$AD=3$，$AB=5$.
（1）求$\frac{AG}{AF}$的值；
（2）求$△ ADE$与$△ ABC$的周长之比；
（3）若$△ ADE$的面积为4，求$△ ABC$的面积.
（1）求$\frac{AG}{AF}$的值；
（2）求$△ ADE$与$△ ABC$的周长之比；
（3）若$△ ADE$的面积为4，求$△ ABC$的面积.
答案
(1) $\because ∠ BAC = ∠ DAE, ∠ ADE = ∠ B, \therefore △ BAC ∽ △ DAE$. 又 $\because AG、AF$ 分别是 $△ BAC$ 和 $△ DAE$ 的高, $\therefore \frac{AG}{AF} = \frac{AB}{AD} = \frac{5}{3}$
(2) $\because △ BAC ∽ △ DAE, \therefore △ ADE$ 与 $△ ABC$ 的周长之比 $= \frac{AD}{AB} = \frac{3}{5}$
(3) $\because △ BAC ∽ △ DAE, \therefore \frac{S_{△ ADE}}{S_{△ ABC}} = (\frac{AD}{AB})^2 = (\frac{3}{5})^2 = \frac{9}{25}. \because △ ADE$ 的面积为 $4, \therefore △ ABC$ 的面积为 $\frac{25}{9} × 4 = \frac{100}{9}$
(2) $\because △ BAC ∽ △ DAE, \therefore △ ADE$ 与 $△ ABC$ 的周长之比 $= \frac{AD}{AB} = \frac{3}{5}$
(3) $\because △ BAC ∽ △ DAE, \therefore \frac{S_{△ ADE}}{S_{△ ABC}} = (\frac{AD}{AB})^2 = (\frac{3}{5})^2 = \frac{9}{25}. \because △ ADE$ 的面积为 $4, \therefore △ ABC$ 的面积为 $\frac{25}{9} × 4 = \frac{100}{9}$
14. 如图，在△ABC中，AB=AC，延长BC至点D，使CD=BC，连结AD，过点C作BD的垂线CE，交AD于点E，连结BE，交AC于点F.
（1）求证：△ABD ∽ △FCB；
（2）若△FCB的面积为9，BC=6，求DE的长.
（1）求证：△ABD ∽ △FCB；
（2）若△FCB的面积为9，BC=6，求DE的长.
答案
14. (1) $\because AB = AC, \therefore ∠ ABC = ∠ ACB. \because EC ⊥ BD, BC = CD, \therefore BE = DE. \therefore ∠ EBD = ∠ EDB. \therefore △ ABD ∽ △ FCB$
(2) 如图, 作 $AH ⊥ BC$ 于点 H.
$\because △ ABD ∽ △ FCB, \therefore \frac{AB}{FC} = \frac{AD}{FB} = \frac{BD}{CB}. \because CD = BC = 6, \therefore \frac{S_{△ FCB}}{S_{△ ABD}} = (\frac{BC}{DB})^2 = \frac{1}{4}. \because S_{△ FCB}=9, \therefore S_{△ ABD}=36. \because BD=12, \therefore AH=6. \because AB=AC, AH ⊥ BC, \therefore BH=CH=3. \therefore DH=9. \because EC ⊥ DB, \therefore EC // AH. \therefore △ DEC ∽ △ DAH. \therefore \frac{EC}{AH} = \frac{DC}{DH}. \therefore \frac{EC}{6} = \frac{6}{9}. \therefore EC=4. \therefore DE = \sqrt{EC^2 + CD^2} = \sqrt{4^2 + 6^2} = 2\sqrt{13}$