11. 如图,已知$□ ABCD的对角线AC$,$BD相交于点O$,且$\angle 1 = \angle 2$.
(1)求证:$□ ABCD$是菱形;
(2)$F为AD$上一点,连接$BF交AC于点E$,且$AE = AF$,求证:$AO = \frac{1}{2}(AF + AB)$.

(1)求证:$□ ABCD$是菱形;
(2)$F为AD$上一点,连接$BF交AC于点E$,且$AE = AF$,求证:$AO = \frac{1}{2}(AF + AB)$.
答案
(1) 证明:∵ $ □ A B C D $ 中,$ A D // B C $,∴ $ \angle 2 = \angle A C B $.
又 ∵ $ \angle 1 = \angle 2 $,∴ $ \angle 1 = \angle A C B $,
∴ $ A B = B C $,∴ $ □ A B C D $ 是菱形.
(2) 证明:∵ $ □ A B C D $ 中,$ A D // B C $,∴ $ \angle A F E = \angle E B C $.
又 ∵ $ A F = A E $,∴ $ \angle A F E = \angle A E F = \angle B E C $,
∴ $ \angle E B C = \angle B E C $,∴ $ B C = C E $,∴ $ A C = A E + C E = A F + B C = 2 O A $,
∴ $ O A = \frac { 1 } { 2 } ( A F + B C ) $.
又 ∵ $ A B = B C $,∴ $ O A = \frac { 1 } { 2 } ( A F + A B ) $.
又 ∵ $ \angle 1 = \angle 2 $,∴ $ \angle 1 = \angle A C B $,
∴ $ A B = B C $,∴ $ □ A B C D $ 是菱形.
(2) 证明:∵ $ □ A B C D $ 中,$ A D // B C $,∴ $ \angle A F E = \angle E B C $.
又 ∵ $ A F = A E $,∴ $ \angle A F E = \angle A E F = \angle B E C $,
∴ $ \angle E B C = \angle B E C $,∴ $ B C = C E $,∴ $ A C = A E + C E = A F + B C = 2 O A $,
∴ $ O A = \frac { 1 } { 2 } ( A F + B C ) $.
又 ∵ $ A B = B C $,∴ $ O A = \frac { 1 } { 2 } ( A F + A B ) $.
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