In a ||gm ABCD, the bisector of angle A also bisects BC at X. Find AB : AD 1. 2 : 1 2. 1 : 2 3. 3 : 2 4. 2 : 3

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In the given figure, ABCD is a parallelogram. E is a point on DC. If AE bisects $∠$ A and BE bisects $∠$ B, then BC : AB = ___.

In a parallelogram ABCD, the bisectors of $∠ A$ also bisect BC at x, find AB : AD.

A (5, 4, 6), B (1, - 1, 3)

C (4, 3, 2)

∠ B A C

A D

A B C D

∠ A

B C

X

A D = 2 A B

A B C D

- - \to A C

- - \to A B

- - \to A D

\frac{2 π}{3}

15 | - - \to A C | = 3 | - - \to A B | = 5 | - - \to A D |

- - \to B A

- - \to D C

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